a,-2x -(x-17)=34-(-x+25)

-2x-x+17=34+x-25

-3x+17=9+x

-3x-x=9-17

-4x=-8

-->4x=8

x=8:4

x=2

Vậy x=2

b,17-(16x-37)=2x+43

17-16x+37=2x+43

20-16x=2x+43

-16x-2x=43-20

-18x=23

x=23:(-18)

x=23/-18

Mà x là số nguyên nên --> x thuộc tập rỗng

c,-2x-3.(x-17)=34-2(-x+25)

-2x-3x+51=34-2.(-x)-25

-5x+51=9-(-2).x

-5x+(-2).x=9-51

-7x=-42

7x=42

x=42:7

x=6

Vậy x=6

\(\Leftrightarrow\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(-6x-18\right)\left(8x^2-18\right)=0\)

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

hay \(x\in\left\{-3;\dfrac{3}{2};-\dfrac{3}{2}\right\}\)

<=> 4x²-3x-18=4x²+3x

<=>6x=-18

<=>x=-3

a)    15-(49-4x)=3x-27

<=> 15-49+4x=3x-27

<=> -34+4x-3x=-27

<=> x=-27+34=7

Vậy x=7

b)   70-(x-37)-(-61)=14-2x

<=> 70-x+34+61=14-2x

<=> 120-x+2x=14

<=> x=120-14=106

Vậy x=106

Bài này sử dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right).\)

a) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left(a^2-\left(b-c\right)^2\right)\left(\left(b+c\right)^2-a^2\right)=\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\left(b+c-a\right)\)

b) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)=-12\left(x+3\right)\left(4x^2-9\right)=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right).\)

c)  \(\left(4abcd+\left(a^2+b^2\right)\left(c^2+d^2\right)\right)^2-4\left(cd\left(a^2+b^2\right)+ab\left(c^2+d^2\right)\right)^2\)

\(=\left(4abcd+\left(a^2+b^2\right)\left(c^2+d^2\right)-2cd\left(a^2+b^2\right)-2ab\left(c^2+d^2\right)\right)\times\)

\(\times\left(4abcd+\left(a^2+b^2\right)\left(c^2+d^2\right)+2cd\left(a^2+b^2\right)+2ab\left(c^2+d^2\right)\right)\)

\(=\left(\left(a^2+b^2\right)\left(c-d\right)^2-2ab\left(c-d\right)^2\right)\times\left(\left(a^2+b^2\right)\left(c+d\right)^2+2ab\left(c+d\right)^2\right)\)

\(=\left(c-d\right)^2\cdot\left(a-b\right)^2\cdot\left(a+b\right)^2\cdot\left(c+d\right)^2.\)

\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=n\)

\(=\left(n+3\right)n-4\)

\(=n^2+3n-4\)

\(=n^2-n+4n-4\)

\(=n\left(n-1\right)+4\left(n-1\right)\)

\(=\left(n-1\right)\left(n+4\right)\)

\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)

\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)

Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):

\(\left(12n+1\right)n=6\)

\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)