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a) \(x^2+5y^2-4xy+6y+9=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y=2.\left(-3\right)=-6\\y=-3\end{matrix}\right.\)
Vậy : \(\left(x,y\right)=\left(-6,-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
Đang 3x2+3y2 sao lại ra -3(x+y)2 ?? Phải là +3(x2+y2) chứ :v Không nhớ hằng đẳng thức 1 và 3 à :v với cả 6xy đâu?
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WTF đăng một loạt vầy ai dám làm @@
Mấy bài này trong sách bài tập cx có bài mẫu
tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết
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\(x+y+z=0\Leftrightarrow x^2+y^2+z^2+2xy+2x+2yz=0\)
\(\Leftrightarrow x^2+y^2+z^2=-2xy-2yz-2xz\)
Có:
\(P=\frac{18\left(x^2+y^2+z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{2\left(x^2+y^2+z^2\right)-2xy-2xz-2yz}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{3\left(x^2+y^2+z^2\right)}=6\)
\(\Leftrightarrow x^2+y^2+z^2=-2xy-2yz-2xz\)
\(P=\frac{18\left(x^2+y^2+z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{2\left(x^2+y^2+z^2\right)-2xy-2xz-2yz}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}\)
\(=\frac{18\left(x^2+y^2+z^2\right)}{3\left(x^2+y^2+z^2\right)}=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=16^{503}=\left(2^4\right)^{503}=2^{2012}\)
\(4x-2y+4=4.2^{2011}-2.2^{2012}+4\)
\(=2^{2013}-2^{2013}+4=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề là CMR $x^4-x^3y+x^2y^2-xy^3+y^4> x^2+y^2$ thì đúng hơn bạn ạ.
Lời giải:
Ta có:
$\text{VT}=(x^4+y^4-x^3y-xy^3)+x^2y^2$
$=(x-y)^2(x^2+xy+y^2)+x^2y^2\geq x^2y^2$
Mà:
$x^2y^2=\frac{x^2y^2}{2}+\frac{x^2y^2}{2}> \frac{x^2.2}{2}+\frac{2.y^2}{2}=x^2+y^2$ do $x^2> 2, y^2>2$
Do đó: $\text{VT}> x^2+y^2$ (đpcm)
b) \(2x^2+4y^2+z^2-4xy-2x-2z+5=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-2x+1\right)+\left(z^2-2z+1\right)+3=0\)
....