Lời giải:

a. $9x^2-16-(3x-4)(2x+5)=0$

$\Leftrightarrow [(3x)^2-4^2]-(3x-4)(2x+5)=0$

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

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

$\Leftrightarrow (3x-4)(x-1)=0$

$\Leftrightarrow 3x-4=0$ hoặc $x-1=0$

$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=1$.

b.

$x^2+4x=12$

$\Leftrightarrow x^2+4x-12=0$

$\Leftrightarrow (x^2-2x)+(6x-12)=0$

$\Leftrightarrow x(x-2)+6(x-2)=0$

$\Leftrightarrow (x-2)(x+6)=0$

$\Leftrightarrow x-2=0$ hoặc $x+6=0$

$\Leftrightarrow x=2$ hoặc $x=-6$

c.

$x^2-2x=35$

$\Leftrightarrow x^2-2x-35=0$

$\Leftrightarrow (x^2+5x)-(7x+35)=0$

$\Leftrightarrow x(x+5)-7(x+5)=0$

$\Leftrightarrow (x+5)(x-7)=0$

$\Leftrightarrow x+5=0$ hoặc $x-7=0$

$\Leftrightarrow x=-5$ hoặc $x=7$

Ta có: \(\frac{x+2}{3}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2\left(x+2\right)}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2x+4}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau được:

\(\frac{2x+4-\left(y-1\right)+z+5}{6-4+7}=\frac{2x+4-y+1+z+5}{6-4+7}=\frac{\left(2x-y+z\right)+\left(4+1+5\right)}{6-4+7}\)

                                                                                                     \(=\frac{17+10}{9}=\frac{27}{9}=3\)

Suy ra: \(2x+4=6.3\Rightarrow2x=14\Rightarrow x=7\)

            \(y-1=3.4\Rightarrow y=13\)

             \(z+5=3.7\Rightarrow z=16\)

Vậy x = 7 ; y = 13; z = 16

b) 2(x+1)+4²=2⁴

      2(x+1)+16=16

    2(x+1) = 16-16

    2(x+1) = 0

       x+1=2x0

       x+1=0

       x=0-1

       x = -1

a, => 2x+5-x+7=18

=>x+12=18

=>x=6

b, =>2(x+1)=0

=>x+1=0

=>x=-1

c, \(\frac{x-3}{x+5}=\frac{5}{7}\Rightarrow7\left(x-3\right)=5\left(x+5\right)\Rightarrow7x-21=5x+25\Rightarrow7x-5x=25+21\Rightarrow2x=46\Rightarrow x=23\)

2x + 5 - ( x - 7 ) = 18

2x + 5 - x + 7 = 18

x = 18 -12=6

2( x + 1 ) +4² = 2⁴

2x + 2 =0

x = -1

\(x-\frac{3}{x}+5=\frac{5}{7}\)  

\(\frac{x^2-3}{x}=\frac{-30}{7}\)

\(7x^2-21=-30x\)

\(7x^2+30x-21=0\)

\(x=0,612471646\)

a)\(\left(x+8\right)-11=20-15\)

\(\left(x+8\right)-11=5\)

\( x+8=5+11\)

\(x+8=16\)

\(x=8\)

b) \(2x-\left(3+x\right)=5-7\)

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

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

\(x=1\)

c) \( \left(3x-2^4\right)\times7^5=2\times7^6\)

\(3x-2^4=2\times\left(7^6:7^5\right) \)

\(\left(3x-2^4\right)=2\times7^2\)

\(3x-2^4=2\times49\)

\(3x-16=98\)

\(3x=114\)

\(x=38\)

a)\(x-15\%x=\frac{1}{3}\)

\(x.\left(1-15\%\right)=\frac{1}{3}\)

\(x.\frac{-280}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{-280}{3}\)

\(x=\frac{-1}{280}\)

Vậy \(x=\frac{-1}{280}\)

b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)

\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)

\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)

\(-\frac{17}{10}x=\frac{-61}{30}\)

\(x=\frac{-61}{30}:\frac{-17}{10}\)

\(x=\frac{61}{51}\)

Vậy \(x=\frac{61}{51}\)