can gap:
giai hpt
x2+1+y(y+x)=4y
(x^2+1)(y+x-2)=-2
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bai 1. Tìm x,y sao cho
a, (3x2+1)2+2xy+y2+1=0
b,x2+2xy+4y2+4y+y2+1=0
cac ban oi giup mih. minh dang can
a, (3x2+1)2+2xy+y2+1=0
(3x2+1)2+(y+1)2=0 Vì (3x2+1)2 >=0 ; (y+1)2 >=0 với mọi x,ý
=>3x2+1=0 => 3x2=1 => x2=1/3 => x=căn 1/3
y+1=0 => y=-1
b, x2+2xy+4y2+4y+y2+1=0
(x2+2xy+y2) + (4y2+4y+1)=0
(x+y)2 + (2y+1)2=0 Vì (x+y)2 >=0 ; (2y+1)2 >=0 vói mọi x,y
=> 2y+1=0 => y=-1/2
x+y=0 => x-1/2=0 => x=1/2
a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
a,
\(\left(x^2-2xy+y^2\right)\left(x-y\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left[\left(x^2-2xy+y^2\right)\left(x-y\right)\right]-\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\)
\(=\left[\left(x-y\right)^2\left(x-y\right)\right]-\left(x-y\right)^3\)
\(=\left(x-y\right)^3-\left(x-y\right)^3\)
\(=0\)
Lời giải:
a.
$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$
$=9x=9.15=135$
b.
$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$
$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$
c.
$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$
$=-8x^3+9x^2y^2-xy^3$
$=(-2x)^3+(3xy)^2-xy^3$
$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$
A = x2(x + y) - y(x2 - y) + 2002
A = x2.x + x2.y + (-y).x2 + (-y)(-y) + 2002
A = x3 + x2y - x2y + y2 + 2002
A = x3 + (x2y - x2y) + y2 + 2002
A = x3 + y2 + 2002 (1)
Thay x = 1, y = -1 vào (1), ta có:
A = x3 + y2 + 2002 = 13 + (-1)2 + 2002
= 1 + 1 + 2002
= 2004
B làm tương tự
a) A = x2( x + y ) - y( x2 + y2 )
= x3 + x2y - x2y - y3
= x3 - y3
Với x = 1 ; y = -1
A = 13 - (-1)3 = 1 + 1 = 2
b) B = 5x( x - 4y ) - 4y( y - 5x )
= 5x2 - 20xy - 4y2 + 20xy
= 5x2 - 4y2
Với x = -0, 6 ; y = -0, 75
B = 5.(-0, 6)2 - 4.(-0, 75)2 = 5.9/25 - 4.9/16 = 9/5 - 9/4 = -9/20
C = x( x - y + 1 ) - y( y + 1 - x )
= x2 - xy + x - y2 - y + xy
= x2 + x - y2 - y
= ( x2 - y2 ) + ( x - y )
= ( x - y )( x + y ) + ( x - y )
= ( x - y )( x + y + 1 )
Thế x = -2/3 ; y = -1/3 ta được
C = [ -2/3 - (-1/3 ) ][ -2/3 - 1/3 + 1 ]
= ( -2/3 + 1/3 ).0
= 0
a, \(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002=x^3-y^3+2002\)
Thay x = 1; y = -1 ta có : \(1^3-\left(-1\right)^3+2002=1-1+2002=2002\)
b, \(5x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}=5x^2-4y^2-\frac{11}{20}\)
Thay x = -0,6 ; y = -0,75 ta có : \(5.\left(-0,6\right)^2-4\left(-0,75\right)^2-\frac{11}{20}=-1\)
c, \(x\left(x-y+1\right)-y\left(y+1-x\right)=x^2+x-y^2-y\)
Thay x = -2/3 ; y = -1/3 ta có : \(\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=0\)