tim cap (x,y ) thoa man
\(x^2.\left(x+3\right)+y^2.\left(y+5\right)-\left(x+y\right).\left(x^2-xy+y^2\right)=0\)
Tìm cap số (x,y) thoa man
\(x^2+y^2=0\)
\(x^2+2y^2+2y\left(1-x\right)=-1\)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(4x^2+y^2-2\left(2x+y-1\right)=0\)
\(2x^2\left(1-y\right)+y\left(y+xy-2x\right)=0\)
\(x^2+y^2=0\)
Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)
(Dấu "="\(\Leftrightarrow x=y=0\))
\(x^2+2y^2+2y\left(1-x\right)=-1\)
\(\Leftrightarrow x^2+2y^2+2y-2xy+1=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y+1\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y+1\right)^2\ge0\)
(Dấu "="\(\Leftrightarrow x=-1;y=-1\)
Cho x,y,z nguyen duong thoa man x+y-z+1=0
Tim GTLN cua \(P=\frac{x^3y^3}{\left(x+yz\right)\left(y+xz\right)\left(z+xy\right)^2}\)
Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)
\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)
\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)
Áp dụng BĐT Cosi ta có:
\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)
\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)
Cho bieu thuc
\(P=\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(x+1\right)}-\frac{x^2y^2}{\left(x+1\right)\left(1-y\right)}\)
Rut gon P= x+xy-y
DKXD \(x\ne y\); \(x\ne-1\):\(y\ne1\)
Tim x y de P nguyen duong thoa man \(x^2+y^2+3xy-x-3y=0\)
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
1. Tim x,y,z biet: \(\frac{1}{2}\left(x+y+z\right)-3=\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-4}\)
2. Chox,y,z > 0 thoa man \(x+y+z+\sqrt{xyz}=4\) . Tinh \(A=\sqrt{x\left(4-y\right)\left(4-z\right)+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}}\)
Cho A = \(\dfrac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}.\left[1:\dfrac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3-y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
B = x - y
Chứng minh đẳng thức A = B
Tính giá trị của A, B tại x = 0; y = 0 và giải thích vì sao A ≠ B
\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
(1)\(\left\{{}\begin{matrix}x^2+y^2=5\\xy\left(x^2-y^2\right)=6\end{matrix}\right.\)
(2)\(\left\{{}\begin{matrix}x^2+y^2+xy=18\\xy\left(x+2y\right)\left(y-x\right)=72\end{matrix}\right.\)
(3)\(\left\{{}\begin{matrix}x\left(x+y+1\right)-3=0\\\left(x+y\right)^2-\dfrac{5}{x^2}+1=0\end{matrix}\right.\)
tim x, y thoa man \(y^2+2\left(x^2+1\right)=2\left(x+1\right)\)
\(y^2+2\left(x^2+1\right)=2\left(x+1\right)\)
\(\Leftrightarrow y^2+2\left(x^2+x+1\right)=2\left(x+1\right)\)
\(\Leftrightarrow y^2+2x^2+2x+2=2x+2\)
\(\Leftrightarrow y^2+2x^2=0\)
Vì \(x^2\ge0;y^2\ge0\)
\(\Rightarrow y^2+2x^2\ge0\)
Mà \(y^2+2x^2=0\)
Nên \(\hept{\begin{cases}y^2=0\\2x^2=0\end{cases}}\)
Hay x = y = 0
cho x,y,z nguyen duong thoa man: \(\left\{{}\begin{matrix}\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\\\left|y-2x\right|\le\dfrac{1}{\sqrt{y}}\end{matrix}\right.\)
tim Max \(A=x^2+2y^2\)
Sau vài phút cố gắng thì khẳng định đề bài của em bị sai