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: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)

Ói , hoa mắt chóng mặt nhức đầu ,

thank you bạn nhiều nha

thank

a) = x² -4x + 4+1 = (x-2)² +1

GYNN = 1

B) = (x² - 8x +16) + ( y² +4y +4) + 2016 -16 -4

     = (x - 4)² + ( y +2)² + 1996

GTNN = 1996

c) =(x-y)² +xy -x +y +1 hình như câu này bn chép sai đề nên k có GTNN

( ĐÂY là những bài khó , có thể bn k hiu nhưng đúng 100%, bn cứ chép ra lop thầy giảng cho)

câu c) có thể là y-1 mà bn nhầm y+1

\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)

\(=2x^2-2xy-y^2+2xy\)

\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)

\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(=5x^2-20xy-4y^2+20xy\)

\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)

\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)

\(=-xy\left(x+1\right)\)

\(=\dfrac{1}{2}.100\left(100+1\right)=50.101=5050\)

cái đoạn\(-xy\left(x+1\right)\)đổi x+1 thành y+1 nha mik đánh nhầm

giúp mik vs mik k cho

mai mik kt 1 tiết r

a,

\(\left(x^2-2xy+y^2\right)\left(x-y\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=\left[\left(x^2-2xy+y^2\right)\left(x-y\right)\right]-\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\)

\(=\left[\left(x-y\right)^2\left(x-y\right)\right]-\left(x-y\right)^3\)

\(=\left(x-y\right)^3-\left(x-y\right)^3\)

\(=0\)

giúp mik vs please

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

cam on bn nhieu nha

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 : \(\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=0\)

a) \(A=x^2\left(x+y\right)-y\left(x^2+y^2+2002\right)\)với \(x=1;y=-1\)

\(A=x^3-y^3-2002y\)

Thay: x=1;y=-1, ta có:

\(1^3-\left(-1\right)^3-2002\left(-1\right)\Rightarrow1+1+2002=2004\)

b) \(B=4x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}\)với \(x=-0,6;y=-0,75\)

\(B=4x^2-4y^2+4xy-\frac{11}{20}\)

\(B=4\left(x^2-y^2+xy\right)-\frac{11}{20}\)

Thay: x= -0,6; y=-0.75, ta có:

\(4\left(-0.6^2--0.75^2+\left(-0,6.-0,75\right)\right)\Rightarrow4\left(0.35.8875\right)=0,143.55\)

c) \(C=x\left(x-y+1\right)-y\left(y+1-x\right)\)với x=-2/3; y=-1/3

\(C=x^2+x-y^2-y\)

Thay: x=-2/3; y=-1/3, ta có:

\(C=\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=\frac{4}{9}-\frac{6}{9}-\frac{1}{9}=-\frac{3}{9}=-\frac{1}{3}\)