f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-4x^2+2x^3-8x+x^2-4=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;1\right\}\)

b) \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)-72=0\)

Đặt \(t=x^2-4\), ta có :

\(t\left(t-6\right)-72=0\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-12=0\\t+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\left(tm\right)\\x^2+2=0\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;-4\right\}\)

c) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc \(x+2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc \(x=-\frac{1}{2}\)

hoặc \(x=-2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)

a, \(x^4+2x^3-3x^2-8x-4=0\)

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

TH1 : \(x+1=0\Leftrightarrow x=-1\)

TH2 : \(x^3+x^2-4x-4=0\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)

=> \(x=-1;x=\pm2\)

b, \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40=72\)

\(\Leftrightarrow x^4-14x^2-32=0\) Đặt \(x^2=t\left(t\ge0\right)\)

Ta có pt mới : \(t^2-14t-32=0\) Tự xử 

Lời giải:
PT \(\Leftrightarrow 2x^4-2x^2+(7x^3-7x)+(3x^2-3)=0\)

\(\Leftrightarrow 2x^2(x^2-1)+7x(x^2-1)+3(x^2-1)=0\)

\(\Leftrightarrow (2x^2+7x+3)(x^2-1)=0\)

\(\Leftrightarrow (2x^2+6x+x+3)(x^2-1)=0\)

\(\Leftrightarrow [2x(x+3)+(x+3)](x^2-1)=0\)

\(\Leftrightarrow (x+3)(2x+1)(x-1)(x+1)=0\Rightarrow \left[\begin{matrix} x=-3\\ x=-\frac{1}{2}\\ x=-1\\ x=1\end{matrix}\right.\)

\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{2x^2+4}{x^2-4}=\frac{2x^2+4}{x^2-4}\)

Vậy phương trình này có vô số nghiệm x thỏa mãn trừ x khác 2 và -2

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`

a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)

\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)

\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)

\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)

\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)

b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

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

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

\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)

a.

\(\left(3x-4\right)\left(x+4\right)+\left(5-x\right)\left(2x^2+3x-1\right)\\ =\left(3x^2+8x-16\right)+\left(10x^2+15x-5-2x^3-3x^2+x\right)\\ =\left(3x^2+8x-16\right)-\left(-2x^3+7x^2+16x-5\right)\\ =2x^3-4x^2-8x-11\)

b.

TT

\(a,\left(3x-4\right)\left(x+4\right)+\left(5-x\right)\left(2x^2+3x+1\right)\)

\(=3x^2+12x-4x-16+10x^2+15x-5-2x^3-3x^2+x\)

\(=-2x^3+10x^2+24x-21\)

\(b,7x\left(x-4\right)-\left(7x-3\right)\left(2x^2-x+4\right)\)

\(=7x^2-28x-\left(14x^3-7x^2+28x+6x^2-3x+12\right)\)

\(=7x^2-28x-\left(14x^3-x^2+25x+12\right)\)

\(=7x^2-28x-14x^3+x^2-25x-12\)

\(=-14x^3+8x^2-53x-12\)