Mình hướng dẫn cách làm chung nhé

f(x) chia hết cho g(x) ⇔ f(x) nhận các nghiệm của g(x) làm nghiệm 

Từ đây dễ rồi :]>

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a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

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

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Rightarrow x=2\)

g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x=14\)

\(\Rightarrow x=\frac{7}{13}\)

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

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

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`

(x + 2)(x - 2) - (x - 2)(x + 5)

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

= (x - 2)-3

= -3x + 6

b) 2x(3x²y + 4x²y - 3)

= 2x(7x²y - 3)

= 14x³y - 6x

bạn giải hết đc ko ạ

Bài 5 : 

f, bạn xem lại đề hay là tìm x chứa tham số a ? 

g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

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

h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)

\(\Leftrightarrow\left(5-x\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=5\)

m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)

n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

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

x=7 nha

a) (x+2)(x-2) - (x-2)(x+5 )

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

= -3 (x-2)

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

= (3x+1 - 1 +2x) (3x+1 +1-2x)

= 5x (x +2)

d) \(x^2\) - 4 - \(\left(x+2\right)^2\)

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

= (x-2) (x+2) - (x+2) (x+2)

= (x+2) (x-2 - x-2)

= -4 (x+2)

e: \(=x^2-16-2x^2-6x+x^2+6x+9=-7\)

b: \(=\left(6x+1-6x+1\right)^2=2^2=4\)

a) \(2^3:\left|x-2\right|=2\)

\(\Leftrightarrow8:\left|x-2\right|=2\)

\(\Leftrightarrow\left|x-2\right|=8:2\)

\(\Leftrightarrow\left|x-2\right|=4\)

Xét trường hợp 1: \(x-2=4\)

\(\Rightarrow x=4+2\)

\(\Rightarrow x=6\)

Xét trường hợp 2: \(x-2=-4\)

\(\Rightarrow x=-4+2\)

\(\Rightarrow x=-\left(4-2\right)\)

\(\Rightarrow x=-2\)

Vậy \(x=6\) hoặc \(x=-2\)

b)