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\(x^2+y^2=0\)
Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)
(Dấu "="\(\Leftrightarrow x=y=0\))
a) \(x^2-8x+y^2+6y+25=0\)
\(\left(x-8\right)x+y\left(y+6\right)+25=0\)
\(x^2+y^2+6y+25=8x\)
\(\Rightarrow x=4,y=-3\)
b ) 4x2-4x+9y2 -12y +5
<=> [( 2x )2 - 4x + 1 ] [ (3y) 2 - 12y + 4 )] = 0
<=> ( 2x - 1 )2 + ( 3y - 2 )2 =0 ( Vì (2x -1)2 >=0 , ( 3y - 2 )2 >= 0 )
<=> 2x - 1 = 0 và 3y -2 = 0
<=> x = 1/2 và y = 2/3
Lời giải:
a)
\(A=\frac{x^2y(y-x)-xy^2(x-y)}{3y^2-2x^2}=\frac{x^2y(y-x)+xy^2(y-x)}{3y^2-2x^2}=\frac{(xy^2+x^2y)(y-x)}{3y^2-2x^2}\)
\(=\frac{xy(x+y)(y-x)}{3y^2-2x^2}=\frac{xy(y^2-x^2)}{3y^2-2x^2}\)
Với $x=-3; y=\frac{1}{2}$ thì:
$xy=\frac{-3}{2}; x^2=9; y^2=\frac{1}{4}$
Do đó $A=\frac{-35}{46}$
b)
\(B=\frac{(8x^3-y^3)(4x^2-y^2)}{(2x+y)(4x^2-4xy+y^2)}=\frac{(2x-y)(4x^2+2xy+y^2)(2x-y)(2x+y)}{(2x+y)(2x-y)^2}\)
\(=4x^2+2xy+y^2=4.2^2+2.2.\frac{-1}{2}+(\frac{-1}{2})^2=\frac{57}{4}\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
1: a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37\) (Vì \(x-y=7\))
\(=100\)
Vậy \(A=100\)
b) Ta có: \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
Vậy \(B=25\)
c) Ta có : \(C=\left(x-y\right)^2\)
\(=x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)-2xy\)
\(=26-2.5\) (Vì \(x^2+y^2=26\) ; \(xy=5\))
\(=16\)
Vậy \(C=16\)
2: a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2\)
\(=x^2+2xy\)
\(=x\left(x+2y\right)\) \(\left(dpcm\right)\)
b) \(\left(x^2+y^2\right)^2-2xy^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\) \(\left(dpcm\right)\)
c) \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)+4xy\)
\(=\left(x-y\right)^2+4xy\) \(\left(dpcm\right)\)
Chúc bn học tốt ✔✔✔
Xin câu a :3
a) (x + y + 1)2 = 3(x2 + y2) + 1
<=> x2 + y2 + 1 + 2xy + 2x + 2y = 3x2 + 3y2 + 1
<=> 2x2 + 2y2 - 2xy - 2x - 2y = 0
<=> (x2 - 2xy + y2) + (x2 - 2x + 1) + (y2 - 2y + 1) = 2
<=> (x - y)2 + (x - 1)2 + (y - 1)2 = 2
Vì 2 = 02 + 12 + 12 nên ta có các TH sau:
TH1:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=1\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=0\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=0\\x=1;y=2\end{matrix}\right.\)
TH3:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=1\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=0;y=1\end{matrix}\right.\)
Vậy ...
a) ta có : \(\left(x+y+1\right)^2=3\left(x^2+y^2\right)+1\)
\(\Leftrightarrow x^2+y^2+1+2xy+2y+2x=3x^2+3y^2+1\)
\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2=\left(x-y\right)^2-2\le0\)
\(\Leftrightarrow-\sqrt{2}\le x-y\le\sqrt{2}\) --> ...
b) \(\left(2x-y-2\right)^2=7\left(x-2y-y^2-1\right)\)
\(\Leftrightarrow4x^2+y^2+4-4xy+4y-4x=7x-14y-7y^2-7\)
\(\Leftrightarrow2x^2-4xy+2y^2+2x^2-11x+\dfrac{121}{16}+6y^2+18y+\dfrac{9}{4}=\dfrac{-19}{16}\left(vl\right)\)
câu c tương tự .