đầu bài là tìm x ạ

\(a,< =>2\left(x-4\right)^2+4x\left(x-4\right)=0< =>\left(x-4\right)\left(2x-8+4x\right)=0\)\(< =>\left(x-4\right)\left(6x-8\right)=0< =>\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

b,\(< =>4x\left(x-2\right)=0< =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c,\(< =>3x\left(x+2\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d,\(< =>4x^2\left(2+x\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

\(a/\)

\(4x-4y+x^2-2xy+y^2\)

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

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

\(=\left(x-y\right)\left(4+x-y\right)\)

\(b/\)

\(x^4-4x^3-8x^2+8x\)

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

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

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

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

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

\(d/\)

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

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

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

\(e/\)(Xem lại đề)

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

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

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

\(=\left(x+1\right)\left(x^3+x+1\right)\)

\(f/\)

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

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

\(=x\left(x-2\right)^2-1\)

\(=[\sqrt{x}\left(x-2\right)]^2-1\)

\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)

\(c/\)

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

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

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

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

\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)

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

a) \(4x-4y+x^2-2xy+y^2\)

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

\(=\left(x-y\right)\left(4+x-y\right)\)

b) \(x^4-4x^3-8x^2+8x\)

\(=x^4+2x^3-6x^3-12x^2+4x^2+8x\)

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

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

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

c) \(x^3+x^2-4x-4\)

\(=x^3-2x^2+3x^2-6x+2x-4\)

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

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

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

\(=\left(x-2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)

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

d) \(x^4-x^2+2x-1\)

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

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

\(=\left(x^2\right)^2-\left(x-1\right)^2\)

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

e)Sửa đề \(x^4+x^3+x^2-1\)

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

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

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

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

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

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

a 4x -4y +(x-y)^2

=4(x-y)+(x-y).(x-y)

=(x-y).(4+x-y)

c x^2(x+1)-4(x+1)

(x+1).(x^2-4)

d x^4-(x^2-2x+1)

=x^4-(x-1)^2

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

MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM

Câu b dễ thôi

\(x^4-4x^3-8x^2+8x\)

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

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

Tìm GTNN của biểu thức :

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

Đặt A = \(x^2+2x+4\)

\(\Leftrightarrow A=\left(x^2+2.x.1+1\right)+3\)

\(\Leftrightarrow A=\left(x+1\right)^2+3\)

Ta luôn có : \(\left(x+1\right)^2\ge0\forall x\)

Suy ra : \(\left(x+1\right)^2+3\ge3\forall x\)

Hay A\(\ge3\) với mọi x

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Nên : \(A_{min}=3khix=-1\)

a) x³ - 16x = 0

x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x⁴ -2x³ + 10x² - 20x = 0

x³ (x - 2) + 10x(x - 2) = 0

(x - 2)(x³ + 10x) = 0

=> x - 2 = 0 hoặc x³ + 10x = 0

x = 2 x(x² + 10) = 0

+ TH1: x = 0

+ TH2: x² + 10 = 0

x² = -10 (vô lí)

Vậy x = 2 hoặc x = 0

c) (2x - 3)² = (x + 5)²

(2x)² + 2 . 2x . 3 + 3² = x² + 2.x.5 + 5²

4x² + 12x + 9 = x² + 10x + 25

4x² + 12x - x² - 10x = 25 - 9

3x² + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!

b: \(\Leftrightarrow32x^5+1-32x^5+1=2\)

=>2=2(luôn đúng)

a: \(\Leftrightarrow\left[\left(x-3\right)^2-\left(x+3\right)^2\right]\left[\left(x-3\right)^2+\left(x+3\right)^2\right]+24x^3=216\)

\(\Leftrightarrow-12x\left(2x^2+18\right)+24x^3=216\)

=>-216x=216

hay x=-1