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a) -4x2 + 8x - 4
= - (4x2 - 8x + 4)
= - (2x - 2)2
b) -x52 + 10 x - 5
= - 5(x2 - 2x + 1)
= - 5(x - 1)2
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x=t\)
\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)
\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)
Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\frac{1}{27}x^3+8y^3\)
3x4y2+3x3y2+3xy2+3y2=3x3y2(x+1)+3y2(x+1)3x4y2+3x3y2+3xy2+3y2=3x3y2(x+1)+3y2(x+1)
=(3x3y2+3y2)(x+1)=3y2(x3+1)(x+1)=(3x3y2+3y2)(x+1)=3y2(x3+1)(x+1)
=3y2(x+1)(x2−x+1)(x+1)=3y2(x2−x+1)(x+1)2
chúc bn hc tốt
a ) ( 2x + 1 )2 - 4 ( x + 2 )2 = 9
4x2 + 4x + 1 - 4 ( x2 +4x + 4 ) = 9
4x2 + 4x + 1 - 4x2 -16x -16 = 9
-12x - 15 = 9
-12x = 24
x = -2
b) 3 ( x - 1 )2 - 3x ( x - 5 ) = 1
3 ( x2 - 2x + 1 ) - 3x2 + 15x = 1
3x2 - 6x + 3 - 3x2 + 15x = 1
9x + 3 = 1
9x = -2
x = \(\frac{-2}{9}\)
b: \(\Leftrightarrow2\left(x^2-2x+1\right)-3x^2+5x-1=0\)
\(\Leftrightarrow2x^2-4x+2-3x^2+5x-1=0\)
\(\Leftrightarrow-x^2+x+1=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-1\right)=5\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{5}}{2}\\x_2=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
c: \(\Leftrightarrow x^2+6x+9-1-\left(x^2+8x-4x-32\right)=0\)
\(\Leftrightarrow x^2+6x+8-x^2-4x+32=0\)
=>2x+40=0
hay x=-20
d: \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow7x^2+8x+13-7x^2+63=36\)
=>8x+76=36
hay x=-5
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
a. Ta có : (x + y)[(x - y)2 + xy]
= (x + y)(x2 - 2xy + y2 + xy)
= (x + y)(x2 - xy + y2)
= x3 + y3
b. Ta có : x3 + y3 - xy(x + y)
= x3 + y3 - x2y - xy2
=x2(x - y) + y2(y - x)
= (x - y)(x2 - y2)
= (x - y)2.(x + y) đpcm
c) Ta có (x + y)3 - 3xy(x + y)
= (x + y)[(x + y)2 - 3xy)
= (x + y)(x2 + 2xy + y2 - 3xy)
= (x + y)(x2 - xy + y2) (đpcm)
a) VP = ( x + y )( x2 - 2xy + y2 + xy ) = ( x + y )( x2 - xy + y2 ) = x3 + y3 = VT ( đpcm )
b) VP = ( x + y )( x - y )2 = ( x + y )( x2 - 2xy + y2 ) = x3 - 2x2y + xy2 + x2y - 2xy2 + y3 = x3 + y3 - x2y - xy2 = x3 + y3 - xy( x + y ) = VT ( đpcm )
c) VP = x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 = x3 + y3 = ( x + y )( x2 - xy + y2 ) = VT ( đpcm )
CẢM ƠN BN NHIỀU