a: A=(4x+5)^2-2*(4x+5)(4x-5)+(4x-5)^2

=(4x+5-4x+5)^2

=10^2=100

b:  B=(3x-2)^2*(3x+2)^2-2(2x+3)(2x-3)

=(9x^2-4)^2-2(4x^2-9)

=81x^4-72x^2+16-8x^2+18

=81x^4-80x^2+34

\(a,A=\left(4x-5\right)^2+\left(4x+5\right)^2+2\left(5+4x\right)\left(5-4x\right)\)

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

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

\(=10^2\)

\(=100\)

\(b,B=\left(3x-2\right)^2\left(3x+2\right)^2-2\left(2x+3\right)\left(2x-3\right)\)

\(=\left(9x^2-4\right)^2-2\left(4x^2-9\right)\)

\(=81x^4-72x^2+16-8x^2+18\)

\(=81x^4-80x^2+34\)

#\(Urushi\)

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{7};5\right\}\)

\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)

\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

Giải tiếp tương tự

Sau đó giải tiếp câu còn lại

a, \(\left|2+3x\right|=\left|4x-3\right|\)

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}}\)

Câu b tương tự

c, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Rightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{5}{8}x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}x+\frac{5}{8}x=-\frac{3}{5}+\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}}\)

d, \(\left|\frac{7}{5}x+\frac{3}{2}\right|-\left|\frac{4}{3}-\frac{1}{4}\right|=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|-\frac{13}{12}=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|=\frac{13}{12}\)

Đến đây dễ rồi, tự làm tiếp :)

P/s: Ko chắc, sai ib với t :v

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!