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Bài 1. a) 4x - 3 = 0
⇔ x = \(\dfrac{3}{4}\)
KL.....
b) - x + 2 = 6
⇔ x = - 4
KL...
c) -5 + 4x = 10
⇔ 4x = 15
⇔ x = \(\dfrac{15}{4}\)
KL....
d) 4x - 5 = 6
⇔ 4x = 11
⇔ x = \(\dfrac{11}{4}\)
KL....
h) 1 - 2x = 3
⇔ -2x = 2
⇔ x = -1
KL...
Bài 2. a) ( x - 2)( 4 + 3x ) = 0
⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)
KL......
b) ( 4x - 1)3x = 0
⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)
KL.....
c) ( x - 5)( 1 + 2x) = 0
⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)
KL.....
d) 3x( x + 2) = 0
⇔ x = 0 hoặc x = -2
KL.....
Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0
⇔ x - 10 ≥ 0
⇔ x ≥ 10
b) 3 - 2( 2x + 3) ≤ 9x - 4
⇔ - 4x - 3 ≤ 9x - 4
⇔ 13x ≥1
⇔ x ≥ \(\dfrac{1}{13}\)
dòng thứ tư câu a quên chưa chuyển vế 15-9 rồi kìa phải là 45x=6 mới đúng nha
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
a) 5 - 4x = 3x - 9
\(\Leftrightarrow5-4x-3x+9=0\)
\(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x-4\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
d) \(4-2x=7-x\)
\(\Leftrightarrow4-2x-7+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
e) \(\left(x+4\right) \left(8-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow9x+6-3x-1-10-12x=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
i) \(3x-6+x=9-x\)
\(\Leftrightarrow3x-6+x-9+x=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
k)\(2t-3+5t=4t+12\)
\(\Leftrightarrow2t-3+5t-4t-12=0\)
\(\Leftrightarrow3t-15=0\)
\(\Leftrightarrow t=5\)
Vậy \(S=\left\{5\right\}\)
đề là gì
a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)
vậy x=2/3 hoặc x=-6
a, (3x-2) (x+6) (x^2 +5) = 0
<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x2 + 5 = 0 (loại vì x2 \(\ge\)0 => x2 + 5 > 0)
<=> x = 2/3 hoặc x = -6
b, (2x+5)^2 = (3x-1)^2
<=> (2x + 5)2 - (3x - 1)2 = 0
<=> (2x + 5 - 3x + 1)(2x + 5 + 3x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)
c, 4x2 (x-1) - x+1 = 0
<=> 4x2(x - 1) - (x - 1) = 0
<=> (x - 1)(4x2 - 1) = 0
<=> (x - 1)(2x - 1)(2x + 1) = 0
vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0
hay x = 1 hoặc x = 1/2 hoặc x = -1/2
a) \(\left(x+6\right)^2-x\left(x+9\right)=0\)
\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)
\(\Leftrightarrow\)\(3x+36=0\)
\(\Leftrightarrow\)\(x=-12\)
Vậy...
b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)
\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)
\(\Leftrightarrow\)\(23x+12=9\)
\(\Leftrightarrow\)\(x=-\frac{3}{23}\)
Vậy
c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)
\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)
\(\Leftrightarrow\)\(16x^2+2x-14=0\)
\(\Leftrightarrow\)\(8x^2+x-7=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)
Vậy
d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)
\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)
\(\Leftrightarrow\)\(-12x+16=0\)
\(\Leftrightarrow\)\(x=\frac{4}{3}\)
Vậy
e) \(\left(x-2\right)^2-\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x^2-4x+4-2x^2+x+6=0\)
\(\Leftrightarrow\)\(-x^2-3x+10=0\)
\(\Leftrightarrow\)\(\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
Vậy
a) (x + 6)(3x + 1) + x2 - 36 = 0
<=> 3x2 + x + 18x + 6 + x2 - 36 = 0
<=> 4x2 + 19x - 30 = 0
<=> 4x2 + 24x - 5x - 30 = 0
<=> 4x(x + 6) - 5(x + 6) = 0
<=> (x + 6)(4x - 5) = 0
<=> x + 6 = 0 hoặc 4x - 5 = 0
<=> x = -6 hoặc x = 5/4
Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!
a) 4x(x - 5) - (x - 1)(4x - 3) = 5
4x2 - 20x - (4x2 - 3x - 4x + 3) = 5
4x2 - 20x - 4x2 + 3x + 4x - 3 = 5
-13x - 3 = 5
\(\Rightarrow\) -13x = 8
\(\Rightarrow\) x = \(\dfrac{-8}{13}\)
b) (3x - 4)(x - 2) = 3x(x - 9) - 3
3x2 - 6x - 4x + 8 = 3x2 - 27x - 3
3x2 - 10x + 8 - 3x2 + 27x + 3 = 0
17x + 11 = 0
\(\Rightarrow\) 17x = -11
\(\Rightarrow\) x = \(\dfrac{-11}{17}\)
c) x2 - 81 = 0
\(\Rightarrow\) x2 = 81
\(\Rightarrow\) x = \(\pm\) 9
d) 3x2 - 75 = 0
3(x2 - 25) = 0
\(\Rightarrow\) x2 - 25 = 0
\(\Rightarrow\) x2 = 25
\(\Rightarrow\) x = \(\pm\)5
e) x2 - 4x + 3 = 0
x2 - x - 3x + 3 = 0
(x2 - x) - (3x - 3) = 0
x(x - 1) - 3(x - 1) = 0
(x - 3)(x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
xin lỗi vì chữa đề