Tìm x:

a) \(3x\left(3x-8\right)-9x^2+8=0\)

b)\(6x-15-x\left(5-2x\right)=0\)

c)  \(x^3-16x=0\)

d) \(2x^2+3x-5=0\)

e) \(3x^2-x\left(3x-6\right)=36\)

f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)

g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)

h) \(4x\left(x-1000\right)-x+1000=0\)

i) \(x^2-36=0\)

j) \(x^2y-2+x+x^2-2y+xy=0\)

k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)

l) \(3x^3-27x=0\)

Tìm số nguyên x

1/ \(17x+3\left(-16x-37\right)=2x+43-4x\)

2/ \(-2x-3\left(x-17\right)=34-2\left(-x+25\right)\)

3/ \(\left\{-3x+2\left[45-x-3\left(3x+7\right)-2x\right]+4x\right\}=55-103-57:\left[-2\left(2x-1\right)^2-\left(-9\right)^0\right]=-106\)

4/ \(-2x+3\left\{12-2\left[3x-\left(20+2x\right)-4x\right]+1\right\}=45\)

5/ \(3x-32>-5+1\)

6/ \(15+4x< 2x-145\)

7/ \(-3\left(2x+5\right)-16< -4\left(3-2x\right)\)

8/ \(-2x+15< 3x-7< 19-x\)

tìm x biết

a) \(\left(x-2\right)^3\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

b)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

c)\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

d)\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

Giúp mk vs đc k ạ mk đg cần gấp 

Ai đúng và nhanh 3 tick nha

Bài 5 Tìm x biết :

a) \(16x^2-\left(4x-5\right)^2=15\)           b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)     d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)                        f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

Bài 6 : Chứng minh đẳng thức  :    \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)

giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)

\(x=0\)

\(x^3=0\)

\(x^3=2.0.\sqrt{0}\)

\(x^3=2x\sqrt{x}\)

\(x^3=2x\sqrt{x}\)     

\(4\left(x^3-2x\sqrt{x}\right)^2=0\)

\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)

\(4x^6-16x^4\sqrt{x}+16x^2x=0\)

\(4x^6+16x^3=16x^4\sqrt{x}\)

\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)

\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)

\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)

\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)

\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)

\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)

\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)

vậy x=0 là nghiệm của pt  =))