a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)

\(\Leftrightarrow x=25\)

b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x=1\)

hay \(x=-\dfrac{1}{9}\)

a: ta có: $\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15$

$\iff 2x^2+4x-5x-10-2x^2+2x=15$

$\iff x=25$

b: Ta có: $\left(5-2x\right)\left(2x+7\right)=4x^2-25$

$\iff 4x^2-25+\left(2x-5\right)\left(2x+7\right)=0$

$\iff \left(2x-5\right)(2x+5+2x+$

\(\Leftrightarrow5x^2-20x+3x-12-x^2+5x=4x^2-25\)

\(\Leftrightarrow-18x=-13\)

hay x=13/18

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

\(C=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

Ta có \(x\left(4x+2\right)-\left(2x-5\right)\left(2x+5\right)=1\)

\(\Leftrightarrow4x^2+2x-\left(4x^2-25\right)=1\)

\(\Leftrightarrow4x^2+2x-4x^2+25=1\)

\(\Leftrightarrow2x=-24\)

\(\Leftrightarrow x=-12\)

Vậy x=-12

\(x\left(4x+2\right)-\left(2x-5\right)\left(2x+5\right)=1\)

\(4x^2+2x-\left(\left(2x\right)^2-5^2\right)=1\)

\(4x^2+2x-4x^2+25=1\)

\(2x+25=1\)

\(2x=-24\)

\(x=-12\)