a) x²+2x+1= (x+1)²

b) 9x²+6xy+y²= (3x+y)²

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

b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)

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Bài 8:

b. 1+8x⁶y³ = 1³+2³(x²)³y³ = 1³+(2x²y)³

= (1+2x²y)(1-2x²y+4x⁴y²)

e. 27x³+\(\dfrac{y^3}{8}\)\(=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)

= (3x+\(\dfrac{y}{2}\))(9x²-\(\dfrac{3xy}{2}\)+\(\dfrac{y^2}{4}\))

Bài 9:

c. 1- 9x +27x² -27x³ = 1³-3.1².3x+3.(3x)²-(3x)³

= (1-3x)³

d. x³+\(\dfrac{3}{2}x^2\)+\(\dfrac{3}{4}x+\dfrac{1}{8}\) = x³+\(3x^2.\dfrac{1}{2}\)+\(3x.\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3\)

= (x+\(\dfrac{1}{2}\))³

f. x² - 2xy +y² -4m² +4m.n - n² = (x² - 2xy +y²)-((2m)² -2.2m.n + n²)

= (x-y)²-(2m-n)² = (x-y-2m+n)(x-y+2m-n)

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

gắp cái gì

Lời giải:

a) $x^3+3x^2y+x+3xy^2+y+y^3$

$=(x^3+3x^2y+3xy^2+y^3)+(x+y)$

$=(x+y)^3+(x+y)=(x+y)[(x+y)^2+1]$

b) $x^3+y(1-3x^2)+x(3y^2-1)-y^3$

$=(x^3-3x^2y+3xy^2-y^3)-(x-y)$
$=(x-y)^3-(x-y)=(x-y)[(x-y)^2-1]=(x-y)(x-y-1)(x-y+1)$

c)

$27x^3+27x^2+9x+1=(3x+1)^3$

d)

$x(x+1)^2+x(x-5)-5(x+1)^2$

$=x(x+1)^2-5(x+1)^2+x(x-5)$
$=(x-5)(x+1)^2+x(x-5)=(x-5)[(x+1)^2+x]$

$=(x-5)(x^2+3x+2)=(x-5)(x+1)(x+2)$

a) Ta có: \(x^2+2x+1\)

\(=x^2+2\cdot x\cdot1+1^2\)

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

b) Ta có: \(1-2y+y^2\)

\(=y^2-2\cdot y\cdot1+1^2\)

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

c) Ta có: \(x^3-3x^2+3x-1\)

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

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

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

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

d) Ta có: \(27+27x+9x^2+x^3\)

\(=x^3+3x^2+6x^2+18x+9x+27\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+9\right)\)

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

e) Ta có: \(8-125x^3\)

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

\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)

f) Ta có: \(64x^3+\frac{1}{8}\)

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

\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)

g) Ta có: \(1-x^2y^4\)

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

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

a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)

b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)

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

d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)

e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)

f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)

Ko chắc ạ!

tự làm đi dễ vầy

Tui cũng tự làm chứ không làm đâu mà nói nhưng có vài câu chia thấy sai nên mới hỏi để xem đáp án của mình có đúng hay không .

a) 5x-15y=5x-3.5.y=5(x-3y)

c) 14xy(xy+28x)

d) \(\dfrac{2}{7}\left(3x-1\right)\left(x-1\right)\)

e) (x-1)³

f) (x+y-2x)(x+y+2x)=(y-x)(3x+y)

g) (3x+\(\dfrac{1}{2}\))(9x²+\(\dfrac{3}{2}x\)+\(\dfrac{1}{4}\))

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

2a)

(x+1)(x²+2x)=0

(x+1)x(x+2)=0

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