bạn viết rõ đề câu a;b nhé 

c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)

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

e, \(\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)

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

`<=>(x+3)(x-2x+1)=0`

`<=>(x+3)(-x+1)=0`

`** x+3=0`

`<=>x=-3`

`** -x+1=0`

`<=>x=1`

`x(x-3) - 5 (x-3) = 0`

`<=>(x-3)(x-5)=0`

`** x-3=0`

`<=>x=3`

`** x-5=0`

`<=>x=5`

`3x + 12 = 0`

`<=>3x=-12`

`<=> x=-4`

`2x (x-2) + 5 (x-2) = 0`

`<=>(x-2)(2x+5)=0`

`** x-2=0`

`<=>x=2`

`** 2x+5=0`

`<=> x= -5/2`

3x^2-x = 0

x(3x-1) = 0

TH1: x = 0

TH2: 3x-1 = 0

x= 1/3

b) Ta có: 2x(x – 5) – x(3 + 2x) = 26

      ⇔ 2 – 10x – 3x – 2 =26

      ⇔ - 13x = 26

      ⇔ x = - 2

2x (x-5) - x(3+2x) = 26

2x^2 - 10x - 3x - 2x^2 = 26

-13x = 26

x = -2

Lời giải:
PT $\Leftrightarrow (x^3-2x^2)+(x^2-4)=0$

$\Leftrightarrow x^2(x-2)+(x-2)(x+2)=0$

$\Leftrightarrow (x-2)(x^2+x+2)=0$

$\Rightarrow x-2=0$ hoặc $x^2+x+2=0$
Nếu $x-2=0\Leftrightarrow x=2$ (tm)

Nếu $x^2+x+2=0$
$\Leftrightarrow (x+\frac{1}{2})^2=-\frac{7}{4}<0$ (vô lý)

Vậy pt có nghiệm duy nhất $x=2$

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

=> cái vế  trong ngoặc là 0.

rồi bạn làm tiếp nha

\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)

\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)

\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)

\(\dfrac{3}{4}-x=\dfrac{91}{66}\)

\(x=\dfrac{-83}{132}\)

a: Ta có: \(-\dfrac{12}{25}\cdot\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\Leftrightarrow x+\dfrac{83}{132}=0\)

hay \(x=-\dfrac{83}{132}\)

a) \(5^x:\left(5^2\right)^2=625\)

\(5^x:625=625\)

\(5^x=5^8\)=> x = 8

Mấy câu kia tương tự

d) \(\left(x-1\right)^4-\left(x-1\right)^4\cdot\left(x-1\right)^3=0\)

\(\left(x-1\right)^4\cdot\left[1-\left(x-1\right)^3\right]=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^3=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy,..........

làm rõ ra đi mà

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)

\(\left(4-8x\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)

\(2\left(x-4\right)-3x\left(x-4\right)=0\)

\(\left(2-3x\right)\left(x-4\right)=0\)

\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

a, Thay x = 25, ta tính được A =  10 7

b, Rút gọn được B =  2 x - 3

c, Ta có A.B =  2 - 4 x + 2   =>  2 + 2 ∈ Ư 4 . Từ đó tìm được x = 0, x = 4