Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a) 2x3 + 6xy - x2z - 3yz
= ( 2x3 + 6xy ) - ( x2z + 3yz )
= 2x( x2 + 3y ) - z( x2 + 3y )
= ( x2 + 3y )( 2x - z )
b) x2 - 6xy + 9y2 - 49
= ( x2 - 6xy + 9y2 ) - 49
= ( x - 3y )2 - 72
= ( x - 3y - 7 )( x - 3y + 7 )
c) x3 + 4x2 + 16x + 64
= ( x3 + 4x2 ) + ( 16x + 64 )
= x2( x + 4 ) + 16( x + 4 )
= ( x + 4 )( x2 + 16 )
a) =(2x^3-x^2z)+(6xy-3yz)
=x^2(2x-z)+3y(2x-z)
=(x^2+3y)(2x-z)
b) =(x^2-6xy+9y^2)-7^2
=(x-3y)^2-7^2
=(x-3y+7)(x-3y-7)
c) =(x^3+4x^2)+(16x+64)
=x^2(x+4)+16(x+4)
=(x^2+16)(x+4)
1, <=> \(\left(4x\right)^2-\left(9y\right)^2\)=\(\left(4x-9y\right)\left(4x+9y\right)\)
1) \(16x^2-81.y^2=\left(4x\right)^2-\left(9.y\right)^2=\left(4x-9y\right)\left(4x+9y\right)\)
2) \(\left(5x-3y\right)^2-\left(3x-5y\right)^2=\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)=\left(2x+2y\right).\left(8x-8y\right)\)
\(=16.\left(x+y\right)\left(x-y\right)\)
3)\(4x^2-y^2+4y-4=4x^2-\left(y^2-4y+4\right)=\left(2x\right)^2-\left(y-2\right)^2=\left(2x-y+2\right).\left(2x+y-2\right)\)
4)\(9.\left(x-y\right)^2-16.\left(2x+y\right)^2=3^2.\left(x-y\right)^2-4^2.\left(2x+y\right)^2=\left(3x-3y\right)^2-\left(8x+4y\right)^2\)
\(=\left(3x-3y-8x-4y\right)\left(3x-3y+8x+4y\right)=\left(-5x-7y\right).\left(11x+y\right)\)
a) 16(4x+5)2 - 25(2x+2)2
\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)
\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)
\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)
\(=\left(26x+30\right)\left(6x+10\right)\)
\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-2y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-3y+5\right)\)
\(c,\left(x+1\right)^4-\left(x-1\right)^4\)
\(=\left(x+1\right)^{2^2}-\left(x-1\right)^{2^2}\)
\(=\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\)
\(=\left(x^2+2x+1+x^2-2x+1\right)\left[\left(x+1+x-1\right)\left(x+1-x+1\right)\right]\)
\(=\left(2x^2+2\right)2x.2\)
\(=4x.2\left(x^2+1\right)\)
\(=8x\left(x^2+1\right)\)
Lời giải:
Những bài này sử dụng những hằng đẳng thức đáng nhớ.
Vì $x=-2$ nên $x+2=0$. Ta có:
\(A=(2x-3)^2-(x-3)^3+(4x+1)[(4x)^2-4x.1+1^2]\)
\(=(2x-3)^2-(x-3)^3+(4x)^3+1^3\)
\(=[2(x+2)-7]^2-(x+2-5)^3+8x^3+1\)
\(=(-7)^2-(-5)^3+8.(-2)^3+1=111\)
--------------------
\(B=(3x-y)^3-[x^3+(2y)^3]+(x+3)^2\)
\(=(3.1-2)^3-(1^3+8.2^3)+(1+3)^2=-48\)
----------------
Vì $x=\frac{1}{2}; y=\frac{-1}{2}\Rightarrow x+y=0$
\(C=(x-5y)^2+(2x-3y)^3-(x-y)^3-[(2x)^3+(3y)^3]\)
\(=(x+y-6y)^2+[2(x+y)-5y]^3-(x+y-2y)^3-[8(x^3+y^3)+19y^3]\)
\(=(-6y)^2+(-5y)^3-(-2y)^3-19y^3\)
\(=36y^2-136y^3=36.(\frac{-1}{2})^2-136(\frac{-1}{2})^3=26\)
1. = (2x)\(^2\) - (3y)\(^2\) = ( 2x + 3y ) ( 2x - 3y )
2. = ( 4 x )\(^3\) + 1\(^3\) = ( 4x + 1 ) ( 16x\(^2\) - 4x + 1 )
3. = ( 4x )\(^2\) - ( 3y - 2 )\(^2\) = ( 4x - 3y + 2 ) ( 4x + 3y - 2 )