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b)
Sửa đề: \(125a^3+75a^2+15a+1\)
Ta có: \(125a^3+75a^2+15a+1\)
\(=\left(5a\right)^3+3\cdot\left(5a\right)^2\cdot1+3\cdot5a\cdot1^2+1^3\)
\(=\left(5a+1\right)^3\)
c) Ta có: \(64-96a+48a^2-8a^3\)
\(=-\left(8a^3-48a^2+96a-64\right)\)
\(=-\left[\left(8a^3-64\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(2a-4\right)\left(4a^2+8a+16\right)-48a\left(a-2\right)\right]\)
\(=-\left[\left(a-2\right)\left(8a^2+16a+32-48a\right)\right]\)
\(=-\left(a-2\right)\left(8a^2-32a+32\right)\)
\(=-8\left(a-2\right)\left(a^2-4a+4\right)\)
\(=-8\left(a-2\right)^3\)
a) `64-96a+48a^2-8a^3`
`=-(8a^3-48a^2+96a-64)`
`=-[(2a)^3 - 3.(2a)^2 .4 + 3.2a.4^2 - 4^3]`
`=-(2a-4)^3`
b) `-m^3n^6-8`
`=-(m^3n^6+8)`
`=-[(mn^2)^3+2^3]`
`=-(mn^2+2)(m^2n^4-2mn^2+4)`.
\(=\left(-5a\right)^3+3.\left(-5a\right)^2+3.\left(-5a\right)+1\)
\(=\left(-5a+1\right)^3\)
a) \(\left(x+2y\right)^2-\left(x-y\right)^2=\left(x+2y+x-y\right)\left(x+2y-x+y\right)\)
\(=\left(2x+y\right).3y\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)
c) \(9x^2-3x+2y-4y^2\)
\(=9x^2-4y^2-3x+2y\)
\(=\left(3x-2y\right)\left(3x+2y\right)-\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left[3x+2y-1\right]\)
d) \(4x^2-4xy+2x-y+y^2\)
\(=4x^2-4xy+y^2+2x-y\)
\(=\left(2x-y\right)^2+2x-y\)
\(=\left(2x-y\right)\left(2x-y+1\right)\)
e) \(x^3+3x^2+3x+1-y^3\)
\(=\left(x+1\right)^3-y^3\)
\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)
g) \(x^3-2x^2y+xy^2-4x\)
\(=x\left(x^2-2xy+y^2\right)-4x\)
\(=x\left(x-y\right)^2-4x\)
\(=x\left[\left(x-y\right)^2-4\right]\)
\(=x\left(x-y+2\right)\left(x-y-2\right)\)
a) (x + 2y)² - (x - y)²
= (x + 2y - x + y)(x + 2y + x - y)
= 3y(2x + y)
b) (x + 1)³ + (x - 1)³
= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]
= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)
= 2x(x² + 3)
c) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) x³ + 3x² + 3x + 1 - y³
= (x³ + 3x² + 3x + 1) - y³
= (x + 1)³ - y³
= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]
= (x - y + 1)(x² + 2x + 1 + xy + y + y²)
g) x³ - 2x²y + xy² - 4x
= x(x² - 2xy + y² - 4)
= x[(x² - 2xy + y²) - 4]
= x[(x - y)² - 2²]
= x(x - y - 2)(x - y + 2)
\(9x^2-4y^2\)
\(=\left(3x\right)^2-\left(2y\right)^2\)
\(=\left(3x-2y\right)\left(3x+2y\right)\)
Bài `1`
\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)
Bài `3`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
a) \(9x^2-16\)
\(=\left(3x\right)^2-4^2\)
\(=\left(3x-4\right)\left(3x+4\right)\)
b) \(x^2+4xy+4y^2-3x-6y\)
\(=\left(x^2+4xy+4y^2\right)-\left(3x+6y\right)\)
\(=\left[x^2+2\cdot x\cdot2y+\left(2y\right)^2\right]-3\left(x+2y\right)\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)
#\(Toru\)
a) Ta có: \(-9x^2+12xy-4y^2\)
\(=-\left(9x^2-12xy+4y^2\right)\)
\(=-\left[\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\right]\)
\(=-\left(3x-2y\right)^2\)
b) Ta có: \(-125a^3+75a^2-15a+1\)
\(=\left(-5a\right)^3+3\cdot\left(-5a\right)^2\cdot1+3\cdot\left(-5a\right)\cdot1^2+1^3\)
\(=\left(-5a+1\right)^3\)
\(=\left(1-5a\right)^3\)
c) Ta có: \(64-96a+48a^2-8a^3\)
\(=4^3-3\cdot4^2\cdot2a+3\cdot4\cdot\left(2a\right)^2-\left(2a\right)^3\)
\(=\left(4-2a\right)^3\)
\(=\left[2\cdot\left(2-a\right)\right]^3\)
\(=8\left(2-a\right)^3\)
d) Ta có: \(-\frac{1}{8}m^3n^6-\frac{1}{27}\)
\(=-\left(\frac{1}{8}m^3n^6+\frac{1}{27}\right)\)
\(=-\left[\left(\frac{1}{2}mn^2\right)^3+\left(\frac{1}{3}\right)^3\right]\)
\(=-\left(\frac{1}{2}mn^2+\frac{1}{3}\right)\left(\frac{1}{4}m^2n^4-\frac{1}{6}mn^2+\frac{1}{9}\right)\)