a) \(=x^2-49-x^2\) \(=-49\)

b) \(=25x^2-1-25x^2-1\) \(=-2\)

c) \(=16x^2-1-16x^2+8x-1\) \(=8x-2\)

d) \(=9x^2-30x+25-9x^2+25\) \(=50-30x\)

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

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

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

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

\(=10^2\)

\(=100\)

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

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

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

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

#\(Urushi\)

\(=705\left(82+19-1\right)=705\cdot100=70500\)

\(=705\left(82+19-1\right)=705.100=70500\)

=70500

\(=\left(4x-5+3x+5\right)^2=\left(7x\right)^2=49x^2\)

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

Bài đã đăng rồi bạn lưu ý không đăng lại làm loãng box toán.

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) \(2\chi-3=3\left(\chi+1\right)\)

\(\Leftrightarrow2\chi-3=3\chi+3\)

\(\Leftrightarrow2\chi-3\chi=3+3\)

\(\Leftrightarrow\chi=-6\)

Vậy phương trình có tập nghiệm S= \(\left\{-6\right\}\)

\(3\chi-3=2\left(\chi+1\right)\)

\(\Leftrightarrow3\chi-3=2\chi+2\)

\(\Leftrightarrow3\chi-2\chi=2+3\)

\(\Leftrightarrow\chi=5\)

Vậy phương trình có tập nghiệm S= \(\left\{5\right\}\)

b) \(\left(3\chi+2\right)\left(4\chi-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+2=0\\4\chi-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-2\\4\chi=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-2}{3}\\\chi=\dfrac{5}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-2}{3};\dfrac{5}{4}\right\}\)

\(\left(3\chi+5\right)\left(4\chi-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+5=0\\4\chi-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-5\\4\chi=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-5}{3}\\\chi=\dfrac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-5}{3};\dfrac{1}{2}\right\}\)

c) \(\left|\chi-7\right|=2\chi+3\)

Trường hợp 1: 

Nếu \(\chi-7\ge0\Leftrightarrow\chi\ge7\)

Khi đó:\(\left|\chi-7\right|=2\chi+3\)

 \(\Leftrightarrow\chi-7=2\chi+3\)

\(\Leftrightarrow\chi-2\chi=3+7\)

\(\Leftrightarrow\chi=-10\) (KTMĐK)

Trường hợp 2:

Nếu \(\chi-7\le0\Leftrightarrow\chi\le7\)

Khi đó: \(\left|\chi-7\right|=2\chi+3\)

\(\Leftrightarrow-\chi+7=2\chi+3\)

\(\Leftrightarrow-\chi-2\chi=3-7\)

\(\Leftrightarrow-3\chi=-4\)

\(\Leftrightarrow\chi=\dfrac{4}{3}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{4}{3}\right\}\)

\(\left|\chi-4\right|=5-3\chi\)

Trường hợp 1:  

Nếu \(\chi-4\ge0\Leftrightarrow\chi\ge4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow\chi-4=5-3\chi\)

\(\Leftrightarrow\chi+3\chi=5+4\)

\(\Leftrightarrow4\chi=9\)

\(\Leftrightarrow\chi=\dfrac{9}{4}\)(KTMĐK)

Trường hợp 2: Nếu \(\chi-4\le0\Leftrightarrow\chi\le4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow-\chi+4=5-3\chi\)

\(\Leftrightarrow-\chi+3\chi=5-4\)

\(\Leftrightarrow2\chi=1\)

\(\Leftrightarrow\chi=\dfrac{1}{2}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)