a. x² = x 

=> x² - x  =0 

=> x(x - 1) = 0

=> x = 0 hoặc x = 1

b. 3x + 12 = 4x + 16

=> 3x + 12 - 4x - 16 = 0

=> (3x - 4x) + (12 - 16) = 0

=> -x - 4 = 0 

=> x = 4

giúp mk với 945 x 239 -1 / 944 + 945 x 238

x²=x

=>x²-x=0

=>x(x-1)=0

=>   x=0. Hoặc x=1

a)(3x-2)(4x-5)=0

3x-2=0 hoặc 4x-5=0

x=\(\frac{2}{3}\) hoặc x=\(\frac{5}{4}\)

b)x(2x-3)-4x+6=0

x(2x-3)-2(2x-3)=0

(2x-3)(x-2)=0

2x-3=0 hoặc x-2=0

x=\(\frac{3}{2}\) hoặc x=2

c)3x(x-1)+2(x-1)=0

(x-1)(3x+2)=0

x-1=0 hoặc 3x+2=0

x=1 hoặc x=\(\frac{-2}{3}\)

d)x²-1=x(x-1)

(x+1)(x-1)=x(x-1)

(x+1)(x-1)-x(x-1)=0

(x-1)(x+1-x)=0

(x-1)1=0

x-1=0

x=1

a) (3x - 2)(4x + 5) = 0

\(\Leftrightarrow\) 3x - 2 = 0 hoặc 4x + 5 = 0

* 3x - 2 = 0 * 4x + 5 = 0

\(\Rightarrow\) 3x = 2 \(\Rightarrow\) 4x = -5

\(\Rightarrow\) x = \(\frac{2}{3}\) \(\Rightarrow x=-\frac{5}{4}\)

vậy x \(\in\) \(\left\{\frac{2}{3};-\frac{5}{4}\right\}\)

b) x (2x - 3) - 4x + 6 = 0

\(\Rightarrow x\left(2x-3\right)-2\left(2x-3\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(x-2\right)=0\)

\(\Rightarrow2x-3=0\) hoặc \(\left(x-2\right)=0\)

* \(2x-3=0\) * \(x-2-0\)

\(\Rightarrow2x=3\) \(\Rightarrow x=2\)

\(\Rightarrow x=\frac{3}{2}\)

vậy \(x\in\left\{\frac{3}{2};2\right\}\)

c) \(3x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Rightarrow\left(3x+2\right)\left(x-1\right)=0\)

\(\Rightarrow3x+2=0\) hoặc \(x-1=0\)

* \(3x+2=0\) * \(x-1=0\)

\(\Rightarrow3x=-2\) \(\Rightarrow x=1\)

\(\Rightarrow x=-\frac{2}{3}\)

vậy pt trên có \(x\in\left\{-\frac{2}{3};1\right\}\)

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a) x=-4

b)4x=3

x=3/4

c)3x=9

x=3

d) 15x=15

x=1

\(3\left(5+x^2\right)-3=2\left(3x-2\right)+16\)

\(\Rightarrow15+3x^2-3=6x-4+16\)

\(\Rightarrow15-16-3+4=6x-3x^2\)

\(\Rightarrow0=3x\left(2-x\right)\)

\(\Rightarrow3x=0\) hoặc 2-x=0

\(\Rightarrow x=0\) hoặc \(x=2\)

a: =>-x+2x=3-7

=>x=-4

b: =>6x+2+2x-5=0

=>8x-3=0

hay x=3/8

c: =>5x+2x-2-4x-7=0

=>3x-9=0

hay x=3

d: =>10x²-10x²-15x=15

=>-15x=15

hay x=-1

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=4\end{matrix}\right.\)

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề