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a, \(4x-3=2\left(x-3\right)\Leftrightarrow4x-3=2x-6\)
\(\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)
b, \(5x^2+x=0\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)
c, \(\left(3x-5\right)\left(x+7\right)=0\Leftrightarrow x=-7;x=\frac{5}{3}\)
d, \(\frac{2}{x-3}-\frac{3}{x+3}=\frac{7x-1}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{2\left(x+3\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{7x-1}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x+6-3x+9=7x-1\Leftrightarrow-x+15=7x-1\)
\(\Leftrightarrow-8x=-16\Leftrightarrow x=2\)( tmđk )
e, \(\left(12x-1\right)\left(6x-1\right)\left(4x-1\right)\left(3x-1\right)=330\)
\(\Leftrightarrow\left(12x-1\right)\left(12x-2\right)\left(12x-3\right)\left(12x-4\right)=330.24=7920\)
\(\Leftrightarrow\left(12x-1\right)\left(12x-4\right)\left(12x-2\right)\left(12x-3\right)=7920\)
\(\Leftrightarrow\left(144x^2-60x+4\right)\left(144x^2-60x+6\right)=7920\)
Đặt \(144x^2-60x+4=t\)
\(t\left(t+2\right)=7920\Leftrightarrow t^2+2t-7920=0\)
\(\Leftrightarrow\left(t-88\right)\left(t+90\right)=0\Leftrightarrow t=88;t=-90\)
suy ra :TH1 : \(144x^2-60x+4=88\Leftrightarrow12\left(12x+7\right)\left(x-1\right)=0\Leftrightarrow x=-\frac{7}{12};x=1\)
TH2 : \(144x^2-60x+4=-90\Leftrightarrow144x^2-60x+94=0\)
\(\Leftrightarrow x=\frac{5\pm3\sqrt{39}i}{24}\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a) \(^{x^3}\) - 7x+6=0
\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0
\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0
\(\Leftrightarrow\) x\(\left(x^2-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!
a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )
<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0
<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0
<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0
<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)
Vậy x = { \(\frac{-1}{3};-5\)}
b) x2 + 10.x + 25 - 4.x . ( x + 5 ) = 0
<=> ( x + 5 )2 -4.x . (x + 5 ) = 0
<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0
<=> ( x + 5 ) . ( 5 - 3.x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)
Vậy x = \(\left\{\frac{5}{3};-5\right\}\)
c) (4.x - 5 )2 - 2. ( 16.x2 -25 ) = 0
<=> ( 4.x-5)2 -2 .( 4.x-5) .( 4.x + 5 ) = 0
<=> ( 4.x -5 )2 - ( 8.x+ 10 ) . ( 4.x -5 ) = 0
<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0
<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0
<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)
Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)
d) ( 4.x + 3 )2 = 4. ( x2 - 2.x + 1 )
<=> 16.x2 + 24.x + 9 - 4.x2 + 8.x - 4 = 0
<=> 12.x2 + 32.x + 5 =0
<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0
<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)
Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
e) x2 -11.x + 28 = 0
<=> x2 -4.x - 7.x + 28 = 0
<=> ( x - 7 ) . ( x - 4 ) = 0
<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
Vậy x = { 4 ; 7 }
f ) 3.x.3 - 3.x2 - 6.x = 0
<=> 3.x. ( x2 -x - 2 ) = 0
<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\([x=0\) \([x=0\)
( Lưu ý :Lưu ý này không cần ghi vào vở : Chị nối 2 ý đó làm 1 nha cj ! )
Vậy x = { 2 ; -1 ; 0 }
Bài 1:
a) 5(x-3)-4=2(x-1)
\(\Leftrightarrow5x-15-4=2x-2\)
\(\Leftrightarrow5x-19-2x+2=0\)
\(\Leftrightarrow3x-17=0\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)
Vậy: \(x=\frac{17}{3}\)
b) 5-(6-x)=4(3-2x)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow-1+x-12+8x=0\)
\(\Leftrightarrow-13+9x=0\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\frac{13}{9}\)
Vậy: \(x=\frac{13}{9}\)
c) (3x+5)(2x+1)=(6x-2)(x-3)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)
\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy:x=1
Bài 2:
a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)
\(\Leftrightarrow4x-10x-15x-3x+60=0\)
\(\Leftrightarrow-24x+60=0\)
\(\Leftrightarrow-24x=-60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy: \(x=\frac{5}{2}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)
\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)
\(\Leftrightarrow-3x=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow3x=-48\)
\(\Leftrightarrow x=-16\)
Vậy: x=-16
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)
\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)
\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)
\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow-13x=-143\)
\(\Leftrightarrow x=11\)
Vậy: x=11
e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)
\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)
\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)
\(\Leftrightarrow45x-18-24-28x+60x-420=0\)
\(\Leftrightarrow77x-462=0\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
Vậy:x=6
Bài 3:
a) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)
Vì \(2\ne0\)
nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)
b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)
c) \(\left(2x+1\right)\left(x^2+2\right)=0\)
Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)
Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta lại có \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)
Ta có: \(4\ne0\)(4)
Từ (3) và (4) suy ra
2x-1=0
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
Bài 4:
a) \(\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)
\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)
\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)
\(\Leftrightarrow-8x^2+40x-32=0\)
\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)
\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)
Vì \(-8\ne0\)
nên \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{1;4\right\}\)
e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)
\(\Leftrightarrow7x^2+58x+115=0\)
\(\Leftrightarrow7x^2+23x+35x+115=0\)
\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)
\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)
Bài 5:
a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)
b) \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)
\(\Leftrightarrow3x^2-3=0\)
\(\Leftrightarrow3\left(x^2-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-1\right\}\)
c) \(x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)
Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)
Từ (5) và (6) suy ra
\(\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy: x=-1
* 4x - 1 = 3x - 2
⇔ 4x - 3x = -2 + 1
⇔ x = -1
Vậy tập nghiệm của pt là S = {-1}
* \(\frac{3}{4}-3x=0\)
⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)
⇒ 3 - 12x = 0
⇔ 12x = 3
⇔ x = \(\frac{3}{12}=\frac{1}{4}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)
* 3x - 2 = 2x + 3
⇔ 3x - 2x = 3 + 2
⇔ x = 5
Vậy tập nghiệm của pt là S = {5}
* 2(x - 3) = 5(x + 4)
⇔ 2x - 6 = 5x + 20
⇔ 2x - 5x = 20 + 6
⇔ -3x = 26
⇔ x = \(\frac{-26}{3}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)
\(A,5x-25=0\)
\(\Leftrightarrow5x-5^2=0\)
\(\Leftrightarrow5\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
Chúc bạn học tốt !