x,y>0 tm \(x+y\ge2\)
CMR \(\sqrt{16x^2y^2+9}\left(\frac{1}{x}+\frac{1}{y}\right)\ge10\)
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Áp dụng BĐT Cô si ta có:
\(x+y\ge2\sqrt{xy}=2\cdot\frac{1}{\sqrt{z}};y+z\ge2\sqrt{yz}=2\cdot\frac{1}{\sqrt{x}};z+x\ge2\sqrt{xz}=2\cdot\frac{1}{\sqrt{y}}.\)( vì xyz=1)
=> P\(\ge\)\(\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}\)+ \(\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(\hept{\begin{cases}a=y\sqrt{y}+2z\sqrt{z}\\b=z\sqrt{z}+2x\sqrt{x}\\c=x\sqrt{x}+2y\sqrt{y}\end{cases}\left(a;b;c\ge0\right)}\)<=> \(\hept{\begin{cases}4a+b=2c+9z\sqrt{z}\\4b+c=2a+9x\sqrt{x}\\4c+a=2b+9y\sqrt{y}\end{cases}}\)
<=> \(\hept{\begin{cases}z\sqrt{z}=\frac{4a+b-2c}{9}\\x\sqrt{x}=\frac{4b+c-2a}{9}\\y\sqrt{y}=\frac{4c+a-2b}{9}\end{cases}}\)
Do đó:
P \(\ge\)\(\frac{2}{9}\cdot\left(\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}+\frac{4c+a-2b}{b}\right)\)
<=> P \(\ge\)\(\frac{2}{9}\left(4\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)+\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)-6\right)\)
<=> P \(\ge\frac{2}{9}\cdot\left(4\cdot3\cdot\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}}+3\cdot\sqrt[3]{\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}}-6\right)\)( Áp dụng BĐT Cô si cho 3 số ko âm)
<=> P \(\ge\frac{2}{9}\left(12+3-6\right)=2\)( đpcm)
Dấu = khi x=y=z=1.
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
Bài toán trở thành \(\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\ge\frac{x+y+z}{3\sqrt{xyz}}\)
Áp dụng bất đẳng thức AM-GM:
\(\frac{z}{x}+\frac{z}{y}+\frac{z}{z}\ge3\sqrt[3]{\frac{z^3}{xyz}}=\frac{3z}{\sqrt[3]{xyz}}\)
Tương tự:
\(\frac{y}{x}+\frac{y}{z}+\frac{y}{y}\ge\frac{3y}{\sqrt[3]{xyz}}\)
\(\frac{x}{z}+\frac{x}{y}+\frac{x}{x}\ge\frac{3x}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow VT+3\ge3+\frac{3}{\sqrt[3]{xyz}}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{3\left(x+y+z\right)}{\sqrt[3]{xyz}}\)\(\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Is it true?
Đặt \(\sqrt{x}=x;\sqrt{y}=y;\sqrt{z}=z\) cho dễ nhìn.
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\x^2+y^2+z^2=2\end{cases}}\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(x\left(1+y^2\right)\left(1+z^2\right)+y\left(1+z^2\right)\left(1+x^2\right)+z\left(1+x^2\right)\left(1+y^2\right)\)
\(=x^2y^2z+y^2z^2x+z^2x^2y+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+x+y+z\)
\(=xyz\left(xy+yz+zx\right)+x^2\left(2-x\right)+y^2\left(2-y\right)+z^2\left(2-z\right)+2\)
\(=-2xyz+2\left(x^2+y^2+z^2\right)-\left(x^3+y^3+z^3-3xyz\right)+2\)
\(=-2xyz+6-\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=-2xyz+6-2=-2xyz+4\)
Ta lại có:
\(\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\)
\(=x^2y^2z^2+\left(xy+yz+zx\right)^2-2xyz\left(xy+yz+zx\right)+3\)
\(=x^2y^2z^2-2xyz+4=\left(xyz-2\right)^2\)
\(\Rightarrow A=\sqrt{\left(xyz-2\right)^2}.\frac{4-2xyz}{\left(xyz-2\right)^2}\)
Tới đây bí :((
e) Sửa đề: \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=2\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)
PT(1) \(\Leftrightarrow x^3+x\left(x-y^2\right)=\sqrt{\left(x-y^2\right)^3}\)
Đặt \(\sqrt{x-y^2}=a.\text{Thay vào, ta có: }x^3+xa^2-2a^3=0\)
Làm tiếp như ở Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học toán với OnlineMath
Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira, Nguyễn Thị Ngọc Thơ, Nguyễn Lê Phước Thịnh, Quân Tạ Minh, An Võ (leo), @tth_new
e nhiều bài quá giải k kịp mn giúp e vs ạ!cần gấp lắm ạ
thanks nhiều!
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)
\(\Leftrightarrow x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}=4\)
\(\Leftrightarrow2+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Khi đó ta có : \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow x+1=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(\Leftrightarrow x+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
Tương tự : \(y+1=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\);
\(z+1=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Ta lần lượt xét các biểu thức :
+) \(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\sqrt{\left[\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\right]^2}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+) \(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
Do đó ta có :
\(P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\cdot\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(P=2\)
Vậy...