a) \(\left(3x+4\right)\left(4-x\right)=0\Rightarrow\orbr{\begin{cases}3x+4=0\\4-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=\left(-4\right)\Rightarrow x=\frac{-4}{3}\\x=4\end{cases}}\)

\(\Rightarrow x=\left\{\frac{-4}{3};4\right\}\)

b) \(\Rightarrow\orbr{\begin{cases}3\left(x-4\right)=0\\2\left(x-4\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-4=0\end{cases}}\Rightarrow x=4\)

c) => 7x²=0+28

=> x²=28:7

=> x²=4

=> x²=2²= (-2)²

=> x={-2;2}

(Ko chép lại đề)

\(a.\Rightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

\(b.\left(3x-2\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=7\end{cases}}\)

\(c.7x^2=28\)

\(x^2=4\)

\(x^2=2^2\)

\(x=\pm2\)

\(d.\left(2x+1\right)\left(1+x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}}\)

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

a)\(\left(3x+5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)

b)\(3x\left(x-7\right)-2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{2}{3}\end{cases}}}\)

c) \(7x^2-28=0\)

\(\Leftrightarrow7x^2=28\)

\(\Leftrightarrow7x^2=7.4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

d)\(\left(2x+1\right)+x\left(2x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}}\)

#H

( 2x - 3 )² - ( x + 5 )² = 0

<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0

<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0

<=> ( x - 8 )( 3x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

7x² - 28 = 0

<=> 7( x² - 4 ) = 0

<=> x² - 4 = 0

<=> x² = 4

<=> x² = (±2)²

<=> x = ±2

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

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

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

\(\Leftrightarrow\left(x-8\right).\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+8\\3x=0-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

Vậy tập hợp nghiệm của phương trình là:  \(S=\left\{8;-\frac{2}{3}\right\}.\)

b) \(7x^2-28=0\)

\(\Leftrightarrow7x^2=0+28\)

\(\Leftrightarrow7x^2=28\)

\(\Leftrightarrow x^2=28:7\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2-4=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+2\\x=0-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy tập hợp nghiệm của phương trình là:  \(S=\left\{2;-2\right\}.\)

a)2x=0+10

2x=10

x=10:2

x=5

b)7x=0+28

7x=28

x=28:7

x=4

c)3x=7+14

3x=21

x=21:3

x=7

d)3(x+2)=44-14

3(x+2)=30

x+2=30:3

x-2=10

x=10+2

x=12

Trường hợp 1: 3x - 0

                    (=) x = 0

TH 2: 4x - 12 = 0

      (=) 4x =12

      (=) x   = 3

TH 3: 7x - 28 = 0

        (=) 7x = 28

tiếp theo

(=) x = 4

Suy ra: x = { 0; 3 ;4}

Tập hợp con của x là

M = { 0 }

N = { 3 }

L = { 4 }

H = tập hợp rỗng

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

(x²-11x+28)(x²-7x+10)-72=0

(x-2)(x-7)(x-4)(x-5)-72=0

(x²-9x+14)(x²-9x+20)-72=0

Đặt x²-9x+14=a, ta có

a(a+6)-72=0

a²+6a-72=0

a²+6a+9-81=0

(a+3)²=81

=> a+3=9 => a=6=> x²-9x+14=6

=>x²-9x+8=0

=> (x-1)(x-8)=0

=> x=1 hoặc x=8

Vậy............ !!!!!

\(\dfrac{4x-5}{7-4x}=\dfrac{7x-28}{26-7x}\)

\(\Leftrightarrow104x-28x^2-130+35x=49x-28x^2-196+112x\)

\(\Leftrightarrow-22x=-66\)

hay x=3

\(\Rightarrow\left(4x-5\right)\left(26-7x\right)=\left(7-4x\right)\left(7x-28\right)\\ \Rightarrow104x-28x^2-130+35x=49x-196-28x^2+112x\\ \Rightarrow-22x=-66\Rightarrow x=3\)