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1: =(4x-1)^2-3(4x-1)
=(4x-1)(4x-1-3)
=4(x-1)(4x-1)
2: =-8x^4y^5(2y+3x)
3: =(a-5)^2-4b^2
=(a-5-2b)(a-5+2b)
5: =x^2-mx-nx+mn
=x(x-m)-n(x-m)
=(x-m)(x-n)
6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)
=(-6a-18)(8a^2-18)
=-6(2a-3)(2x+3)(a+3)
a) = 2x2y2(3y2 - 4y2 + 5y)
= 2x2y2 * ( - y2 +5y)
=2x2y2 * y(5-y)
PTĐTTNT
a)
\(6x^2y^4-8x^2y^2+10x^2y^3\)
\(=x^2y^2\left(6y^2-8+10y\right)\)
b)
\(x^2+y^2-3x-3y+2xy\)
\(=x^2+2xy+y^2-3x-3y\)
\(=\left(x+y\right)^2-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-3\right)\)
c)
\(x^2-25-5+2\sqrt{5}\)
\(=x^2-5^2-5+2\sqrt{5}\)
\(=x^2-5\left(5+1+\sqrt{2}\right)\)
a) 10x(x-y)-6y(y-x)=10x(x-y)+6y(x-y)=(10x+6y)(x-y)
b) \(x^2-25-2xy+y^2=x^2-2xy+y^2-25=\left(x-y\right)^2-25\)
\(=\left(x-y+5\right)\left(x-y-5\right)\)
c) \(x^2-5x+5y-y^2=\left(x^2-y^2\right)-\left(5x-5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)
d)\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)\(=\left(x+3\right)\left(x+1\right)\)
e)\(x^2-4x-5=x^2-5x+x-5=x\left(x-5\right)+\left(x-5\right)\)\(=\left(x+1\right)\left(x-5\right)\)
\(a,2x^2-2xt-5x+5y\)
\(=\left(2x^2-5x\right)-\left(2xy-5y\right)\)
\(=x\left(2x-5\right)-y\left(2x-5\right)\)
\(=\left(2x-5\right)\left(x-y\right)\)
\(b,8x^2+4xy-2ax-ay\)
\(=\left(8x^2-2ax\right)+\left(4xy-ay\right)\)
\(=2x\left(4x-a\right)+y\left(4x-a\right)\)
\(=\left(4x-a\right)\left(2x+y\right)\)
\(c,x^3-4x^2+4x\)
\(=x^3-2x^2-2x^2+4x\)
\(=\left(x^3-2x^2\right)-\left(2x^2-4x\right)\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x-2\right)\)
\(=x\left(x-2\right)^2\)
\(d,2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
\(e,x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+2yz+z^2\right)\)
\(=x^2-\left(y+z\right)^2=\left(x-y-z\right)\left(x+y+z\right)\)
Lời giải:
ĐK: $x\neq 5;x\neq 0; y\neq 2; y\neq -1$
\(M=\frac{x^2-25}{x^3-10x^2+25x}:\frac{y-2}{(y-2)(y+1)}=\frac{(x-5)(x+5)}{x(x^2-10x+25)}:\frac{1}{y+1}\)
\(=\frac{(x-5)(x+5)}{x(x-5)^2}:\frac{1}{y+1}=\frac{x+5}{x(x-5)}.(y+1)=\frac{(x+5)(y+1)}{x(x-5)}\)
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$x^2+9y^2-4xy=2xy-|x-3|$
$\Leftrightarrow x^2+9y^2-6xy=-|x-3|$
$\Leftrightarrow (x-3y)^2+|x-3|=0$
Dễ thấy $(x-3y)^2\geq 0; |x-3|\geq 0$ với mọi $x,y\in $ĐKXĐ nên để tổng của chúng bằng $0$ thì:
$x-3y=x-3=0\Rightarrow x=3; y=1$
Khi đó: $M=\frac{(3+5)(1+1)}{3(3-5)}=\frac{-8}{3}$
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
25-x2+2xy-y2=25-(x-y)2=(5-x+y)(5+x-y)