Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x+y+z=0
nên x+y=-z; y+z=-x; x+z=-y
\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}=-1\)
BĐT: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Nếu ko bạn có thể làm theo AM-GM:
\(\frac{1}{1+x}+\frac{1+x}{4}\ge2\sqrt{\frac{1+x}{4\left(x+1\right)}}=1\)
Tương tự: \(\frac{1}{1+y}+\frac{1+y}{4}\ge1\) ; \(\frac{1}{1+z}+\frac{1+z}{4}\ge1\)
Cộng vế với vế:
\(A+\frac{3+x+y+z}{4}\ge3\Rightarrow A\ge3-\frac{3+x+y+z}{4}\ge3-\frac{3+3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(A=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{9}{1+x+1+y+1+z}=\frac{9}{3+x+y+z}\ge\frac{9}{3+3}=\frac{3}{2}\)
\(A_{min}=\frac{3}{2}\) khi \(x=y=z=1\)
Câu 1:
\(x\left(x-2\right)\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=4\)
\(\Leftrightarrow x\left(x^2-4\right)-\left(x^3+8\right)=4\)
\(\Leftrightarrow x^3-4x-x^3-8=4\)
\(\Leftrightarrow-4x-8=4\)
\(\Leftrightarrow-4x=12\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3\)
a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)
Vậy: Khi x-y=7 thì A=100
b) Ta có: \(x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy+10=4\)
\(\Leftrightarrow2xy=-6\)
\(\Leftrightarrow xy=-3\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)
Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:
\(A=2\cdot\left(10+3\right)=2\cdot13=26\)
Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26
\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)
\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2013}=\frac{1}{x+y+z}\Rightarrow\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\Rightarrow\left(yz+xz+xy\right)\left(x+y+z\right)=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz+xyz=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz=0\)
\(\Rightarrow\left(x^2y+x^2z+xy^2+xyz\right)+\left(y^2z+xz^2+y^2z+xyz\right)=0\)
\(\Rightarrow x\left(xy+xz+y^2+yz\right)+z\left(yz+xz+y^2+xy\right)=0\)
\(\Rightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\Rightarrow x^3+y^3=0\\y+z=0\Rightarrow y^5+z^5=0\\x+z=0\Rightarrow z^7+x^7=0\end{cases}}\)
\(\Rightarrow A=\left(x^3+y^3\right)\left(y^5+z^5\right)\left(z^7+x^7\right)=0\)
\(A=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{9}{3+x+y+z}\ge\frac{9}{3+3}=\frac{3}{2}\)
\(\Rightarrow A_{min}=\frac{3}{2}\) khi \(x=y=z=1\)
Lời giải:
Áp dụng BĐT Cô-si cho các số dương:
\(\frac{1}{x+1}+\frac{x+1}{4}\geq 1; \frac{1}{y+1}+\frac{y+1}{4}\geq 1; \frac{1}{z+1}+\frac{z+1}{4}\geq 1\)
Cộng theo vế:
\(\Rightarrow A+\frac{x+y+z+3}{4}\geq 3\)
\(\Leftrightarrow A\geq \frac{9}{4}-\frac{x+y+z}{4}\)
Mà \(x+y+z\leq 3\Rightarrow \Leftrightarrow A\geq \frac{9}{4}-\frac{x+y+z}{4}\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow x=y=z=1\)
-------------
Hoặc bạn có thể áp dụng luôn BĐT Cauchy-Schwarz:
\(A\geq \frac{(1+1+1)^2}{1+x+1+y+1+z}=\frac{9}{x+y+z+3}\geq \frac{9}{3+3}=\frac{3}{2}\)