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a, 15x3 - 15x = 0
15x(x2-1)=0
15x=0 hoặc x2-1=0 (tự tính nhoa)
b,3x2-6x+3=0
3(x2-2x+1)=0
x2 -2x+1=0:3=3
x2-2x=3-1=2
x(x-2)=0
x=0 hoặc x-2=0 (tự tính nhoa)
Bài làm
a) 15x3-15x=0
<=> 15x( x2 - 1 ) = 0
<=> \(\orbr{\begin{cases}15x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy x = { 0; + 1 }
b) 3x2 - 6x + 3 = 0
<=> 3( x2 - 2x + 1 ) = 0
<=> x2 - 2x + 1 = 0
<=> ( x - 1 )2 = 0
<=> x - 1 = 0
<=> x = 1
Vậy x = 1
c) 5(x - 1) - 3x(1 - x) = 0
<=> 5(x - 1) + 3x(x - 1) = 0
<=> (5 + 3x)(x - 1) = 0
<=> \(\orbr{\begin{cases}5+3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=1\end{cases}}}\)
Vậy x = { -5/3; 1 }
e) -7(x + 2) = 2x(x + 2)
<=> -7(x + 2 ) - 2x( x + 2 ) = 0
<=> (x + 2)(-7 - 2x) = 0
<=> \(\orbr{\begin{cases}x+2=0\\-7-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{2}\end{cases}}}\)
Vậy x = { -2; x = -7/2 }
f)(2x - 3)(3x + 5) = (x - 1)(3x + 5)
<=> (2x - 3)(3x + 5) - (x - 1)(3x + 5) = 0
<=> (3x + 5)(2x - 3 - x + 1) = 0
<=> (3x + 5)(x - 2) = 0
<=> \(\orbr{\begin{cases}3x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=2\end{cases}}}\)
Vậy x = { -5/3; 2 }
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
c,
<=> \(\left[\begin{matrix}x-1=0\\x^2+5x+2=0\\x^3-1=0\end{matrix}\right.\)
+/ x - 1 = 0 <=> x = 1
+/x2 + 5x + 2 =0 <=> (x + \(\frac{5}{2}\))2 - \(\frac{17}{4}\)= 0 <=> (x + \(\frac{5}{2}\))2 = \(\frac{17}{4}\)<=> x + \(\frac{5}{2}\)= \(\pm\)\(\sqrt{\frac{17}{4}}\)
<=> x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
+/ x3 - 1 = 0 <=.> ( x - 1 )(x2 + x + 1 ) = 0
<=> x = 1
Vậy phương trình có Nghiệm là x = 1 và x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
d,
x2 + (x + 3)(10 -2x ) = 9
<=> x2 + 10x - 2x2 + 30 - 6x -9 = 0
<=> x2 + 4x + 21 = 0
<=> 7x - x2 + 21 -3x = 0
<=> (x +3)(7-x) =0
<=> \(\left[\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.\) <=> \(\left[\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy pt có nghiệm là x = -3 và x = 7
a) \(3x^3-6x^2=0\)
\(3x^2\left(x-2\right)=0\)
\(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) \(x\left(x-4\right)-12x+48=0\)
\(x^2-4x-12x+48=0\)
\(x^2-16x+48=0\)
\(\left(x-12\right)\left(x-4\right)=0\)
\(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)
c) Viết thiếu nha :v
d) \(2x\left(x-5\right)-x\left(2x+3\right)=16\)
\(2x^2-10x-x^2-2x^2-3x=16\)
\(-13x=16\)
\(x=-\frac{16}{13}\)
e) \(\left(4x^2-1\right)-\left(x-1\right)^2=-3\)
\(4x^2-1-x^2+2x-1=-3\)
\(3x^2-2+2x=-3\)
\(3x^2-2+2x+3=0\)
\(3x^2+1+2x=0\)
Vì \(3x^2+1+2x>0\)nên:
\(x\in\varnothing\)
A) 3x3 - 6x2 = 0
=> 3x2(x - 2) = 0
=> \(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) x(x - 4) - 12x + 48 = 0
=> x(x - 4) - 12(x - 4) = 0
=> (x - 12)(x - 4) = 0
=> \(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)
c) x(x - 4) - (x2 - 8) = x2 - 4x - x2 + 8 = 4x + 8
a) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\)
\(\Leftrightarrow x=-21\)
Vậy ...
c) \(5x\left(12x+7\right)-3x\left(2x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-6x^2+15x+100=0\)
\(\Leftrightarrow54x^2+50x+100=0\)
\(\Leftrightarrow54\left(x^2+\frac{25}{27}x+\frac{625}{2916}\right)+\frac{290975}{2916}=0\)
\(\Leftrightarrow54\left(x+\frac{25}{54}\right)^2+\frac{290975}{2916}=0\left(ktm\right)\)
Vậy phương trình vô nghiệm.
d) \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x-5=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy ...
e) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)
\(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
\(\Leftrightarrow-2x^2=0\)
\(\Leftrightarrow x=0\)
Vậy ...
a) Ta có: (2x-3)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-2\right\}\)
b) Ta có: (3x-1)(2x-5)=(3x-1)(x+2)
⇔\(\left(3x-1\right)\left(2x-5\right)-\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[\left(2x-5\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{3};7\right\}\)
c) Ta có: \(\left(x^2-25\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5+2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\cdot3\cdot\left(x-2\right)=0\)
mà 3≠0
nên \(\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
Vậy: x∈{5;2}
d) Ta có: \(\left(x^2-6x+9\right)-4=0\)
\(\Leftrightarrow\left(x-3\right)^2-2^2=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy: x∈{5;1}
e) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;\frac{3}{2}\right\}\)
a, \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b, \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x^2=-1\left(voly\right)\end{cases}\Leftrightarrow}x=-1\)
c, \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
d, \(2x\left(3x-5\right)=10-6x\)
\(\Leftrightarrow6x^2-10x-10+6x=0\)
\(\Leftrightarrow\left(6x^2+6x\right)-\left(10x+10\right)=0\)
\(\Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(6x-10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\6x-10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\6x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)