1 , Tìm x,y biết
a 3x2 - 3x(x - 2) = 36
b, x2 + 2xy + 2y2 - 2y + 1 = 0
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bài 1
Ta có 2xy+x+y=83
<=>4xy+2x+2y=166
2x*(2y+1)+(2y+1)=167
(2x+1)*(2y+1)=167
=>2x+1;2y+1 thuộc Ư(167)
do x,y nguyên dương =>2TH
TH1 2x+1=1;2y+1=167=>x=...;y=....
TH2 2x+1=167;2y+1=1=>x=...;y=....
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a) $(2x-1)^2-2(2x-3)^2+4$
$=4x^2-4x+1-2(4x^2-12x+9)+4$
$=4x^2-4x+1-8x^2+24x-18+4$
$=-4x^2+20x-13$
b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$
Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
$=4x^2-4x+1-2(4x^2-12x+9)+4$
$=4x^2-4x+1-8x^2+24x-18+4$
$=-4x^2+20x-13$
b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:
$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$
$=[(3x+2)-(2y-1)]^2$
$=(3x-2y+3)^2$
c) $(x^2+2xy)^2+2(x^2+2xy)y^2+y^4$$=(x^2+2xy)^2+2(x^2+2xy)y^2+(y^2)^2$
$=(x^2+2xy+y^2)^2$
d) $(x-1)^3+3x(x-1)^2+3x^2(x-1)+x^3$
Đặt $a=x-1,\ b=x$:
$=a^3+3a^2b+3ab^2+b^3$
$=(a+b)^3$
$=[(x-1)+x]^3$
$=(2x-1)^3$
e) $(2x+3y)(4x^2-6xy+9y^2)$Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:
$=(2x)^3+(3y)^3$
$=8x^3+27y^3$
f) $(x-y)(x^2+xy+y^2)-(x+y)(x^2-xy+y^2)$$=x^3-y^3-(x^3+y^3)$
$=-2y^3$
g) $(x^2-2y)(x^4+2x^2y+4y^2)-x^3(x-y)(x^2+xy+y^2)+8y^3$Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:
$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$
$=x^6-8y^3-x^6+x^3y^3+8y^3$
$=x^3y^3$
$=[(x+2)-(x+4)][(x+2)+(x+4)]+x^2-3x+1$
$=(-2)(2x+6)+x^2-3x+1$
$=-4x-12+x^2-3x+1$
$A=x^2-7x-11$
b) $(2x+2)^2-4x(x+2)$$=4(x+1)^2-4x(x+2)$
$=4(x^2+2x+1)-4x^2-8x$
$B=4$
a, 3x2 - 3x ( x - 2 ) = 36
3x2 - 3x2 + 6x = 36
6x = 36
=> x=6
\(\Rightarrow x^2+2xy+y^2+y^2-2y+1=0\)
\(\Rightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)