a) x⁴ + 2x² + 1

= (x²)² + 2.x².1 + 1²

= (x² + 1)²

b) 4x² - 12xy + 9y²

= (2x)² - 2.2x.3y + (3y)²

= (2x - 3y)²

c) -x² - 2xy - y²

= -(x² + 2xy + y²)

= -(x + y)²

d) (x + y)² - 2(x + y) + 1

= (x + y)² - 2.(x + y).1 + 1²

= (x - y + 1)²

e) x³ - 3x² + 3x - 1

= x³ - 3.x².1 + 3.x.1² - 1³

= (x - 1)³

g) x³ + 6x² + 12x + 8

= x³ + 3.x².2 + 3.x.2² + 2³

= (x + 2)³

h) x³ + 1 - x² - x

= (x³ + 1) - (x² + x)

= (x + 1)(x² - x + 1) - x(x + 1)

= (x + 1)(x² - x + 1 - x)

= (x + 1)(x² - 2x + 1)

= (x + 1)(x - 1)²

k) (x + y)³ - x³ - y³

= (x + y)³ - (x³ + y³)

= (x + y)³ - (x + y)(x² - xy + y²)

= (x + y)[(x + y)² - x² + xy - y²]

= (x + y)(x² + 2xy + y² - x² + xy - y²)

= (x + y).3xy

= 3xy(x + y)

1)

a) \(=15x^3-20x^2+10x\)

b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)

2) 

a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)

\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))

b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)

(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ

Giúp em vs ạ . Em cảm ơn

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

c) \(=\left(4x-3y\right)\left(4x+3y\right)\)

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)

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

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

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

a: \(C=x^3-3x^2+3x+2023\)

\(C=x^3-3x^2+3x-1+2024\)

\(=\left(x-1\right)^3+2024\)

Khi x=101 thì \(C=\left(101-1\right)^3+2024\)

\(=100^3+2024\)

\(=1000000+2024=1002024\)

b: \(D=x^3-6x^2+12x-100\)

\(=x^3-6x^2+12x-8-92\)

\(=\left(x-2\right)^3-92\)

Khi x=-98 thì \(D=\left(-98-2\right)^3-92\)

\(=-100^3-92\)

\(=-1000000-92=-1000092\)

ai biết chính xác thì giúp mình nhé

ai biet de bai thi giup ban ay nha

a,= (x+4)\(^3\)

b,= (x-2)\(^3\)

c,= x\(^3\)+8

d,=x\(^3\)-27

a,= x\(^3\)+3.4x\(^2\)+3.4\(^2\).x+4\(^3\)=(x+4)\(^3\)

-3.2.x\(^2\)+3.2\(^2\).x-2\(^3\)= (x-2)\(^3\)

còn c,d áp dụng HĐT là ra! ( đc chx bà nội)

a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)

b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))

c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)

a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)

      \(\Rightarrow x=1\)

b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)

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

c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)

     \(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)

     \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)

cdbdb

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...