a) \(16x^2-8x+1=\left(4x\right)^2-2.4x.1+1^2=\left(4x-1\right)^2\)\(27x^3-27x^2+9x-1=\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1^2-1^3=\left(3x-1\right)^3\)c) \(25x^2+20x+4=\left(5x\right)^2+2.5x.2+2^2=\left(5x+2\right)^2\) d) \(x^3+6x^2+12x+8=x^3+3x^2.2+3x.2^2+2^3=\left(x+2\right)^3\)

Viết về bình phương của 1 tổng hoặc 1 hiệu. Lập phương của 1 tổng hoặc 1 hiệu

a)$16x^2-8x+1$

b)$27x^3-27x^2+9x-1$

c) $25x^2+20x+4$

d) $x^3+6x^2+12x+8$

TRẢ LỜI

a/(4x)^2-2.4x+1^2=(4x-1)^2

b/SAI Đề nhé phải là 27x^3-9x^2+27x-1=(3x-1)^3

c/(5x)^2+2.5x+2^2=(5x-2)^2

d/x^3+3.2.x^2+3.2^2.x+2^3=(x+2)^3

a) \(16x^2-8x+1=\left(4x-1\right)^2\)

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

c) \(25x^2+20x+4=\left(5x+2\right)^2\)

d) \(x^3+6x^2+12x+8=\left(x+2\right)^3\)

viết các biểu tức sau dưới dạng lập phương của 1 tổng hoặc lập phương của 1 hiệu

a) \(x^3+12x^2+48x+64\)

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

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

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

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

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

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

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

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

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

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

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

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

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

b) \(48x+64+x^3+12x^2\)

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

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

c) \(-9y^2+y-\dfrac{1}{27}+27y^3\)

\(=27y^3-9y^2+y-\dfrac{1}{27}\)

\(=\left(3y\right)^3-3.\left(3y\right)^2.\dfrac{1}{3}+3.3y.\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3\)

\(=\left(3y-\dfrac{1}{3}\right)^3\)

d) \(8x^3+150x-125-60x^2\)

\(=8x^3-60x^2+150x-125\)

\(=\left(2x\right)^3-3.\left(2x\right)^2.5+3.2x.5^2-5^3\)

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

a, \(8-12x+6x^2-x^3=-\left(x^3-6x^2+12x-8\right)\)

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

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

b, \(48x+64+x^3+12x^2=x^3+4x^2+8x^2+32x+16x+24\)

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

c, \(-9y^2+y-\dfrac{1}{7}+27y^3\)

(sai đề)

d, \(8x^3+150x-125-60x^2=8x^3-20x^2-40x^2+100x+50x-125\)

\(=4x^2\left(2x-5\right)-20x\left(2x-5\right)+25\left(2x-5\right)\)

\(=\left(2x-5\right)\left(4x^2-20x+25\right)=\left(2x-5\right)\left(2x-5\right)^2\)

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

Chúc bạn học tốt!!!

a) Ta có: \(x^3+12x^2+48x+64\)

\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)

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

b) Ta có: \(x^3-12x^2+48x-64\)

\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)

\(=\left(x-4\right)^3\)

c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

d)Sửa đề: \(x^3-3x^2+3x-1\)

Ta có: \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

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

e) Ta có: \(8-12x+6x^2-x^3\)

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

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

f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)

\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)

\(=\left(\frac{1}{3}-3y\right)^3\)

1+6x+12x² + 8x³

=(1+2x)³

y³+3y+3y²+1

=(y+1)³

\(x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3\)

\(x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3\)

a) \(A=8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

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

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

d)  \(D=27+27y^2+9y^4+y^6=\left(3+y^2\right)^3\)

a)-x^3+3x^2-3x+1

=-(x³-3x²+3x-1)

=-(x-1)³

b)8-12x+6x^2-x^3

=2³-3.2².x+3.2.x²-x³

=(2-x)³