\(\left|\frac{2-x}{x+1}\right|\ge2\)
Những câu hỏi liên quan
x;y;z>0. CMR: \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge2+\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
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Tìm x biết:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\left(x\in N,x\ge2\right)\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
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TL:
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right)2x}\right)=\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{4x}=\frac{1}{8}\)
\(\frac{1}{4x}=\frac{3}{8}\)
=>x=2/3
hc tốt
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i dont know
Cho \(x,y,z\) là các số thực dương. CMR: \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge2+\frac{2\left(x+y+z\right)}{3\sqrt{xyz}}\)
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
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?
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\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge2\sqrt{\frac{x^2}{y-1}.\frac{y^2}{x-1}}=2\sqrt{\frac{x^2}{x-1}.\frac{y^2}{y-1}}\ge2\sqrt{4.4}=8\)(cosi)
Vì \(\frac{x^2}{x-1}\ge4;\frac{y^2}{y-1}\ge4\)(vì \(\left(x-2\right)^2\ge0;\left(y-2\right)^2\ge0\))
dấu bằng xảy ra khi x=y=2
- Chứng minh :
a) \(\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{\left(x+2\right)\left(x-2\right)+x+2}}=\frac{1}{\sqrt{x+2}}\left(x>2\right)\)
b) \(\frac{x^2+2}{\sqrt{x^2+1}}\ge2\)
c) \(\frac{2x^2+1}{\sqrt{4x^2+1}}\ge1\)
A frac{x-4sqrt{x}+2}{sqrt{x}-2} left(xge0;xne1right)B frac{xsqrt{x}-1}{x-1} left(xge0;xne1right)C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}}left(xge2right)D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}}left(xge2right)E frac{x+sqrt{x^2-2x}}{x-sqrt{x^2}-2x}-frac{x-sqrt{x^2}-2x}{x+sqrt{x^2}-2x}
Đọc tiếp
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne1\right)\)
B = \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)\(\left(x\ge2\right)\)
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)\(\left(x\ge2\right)\)
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2}-2x}{x+\sqrt{x^2}-2x}\)
cho x,y,z > 0 , x+ y + z\(\le\)1
chứng minh :
\(2\left(x+y+z\right)+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\)
Cho ba số thực x,y,z phân biệt. Chứng minh rằng:
\(\frac{x^2}{\left(y-z\right)^2}+\frac{y^2}{\left(z-x\right)^2}+\frac{z^2}{\left(x-y\right)^2}\ge2\)
Giải các bất phương trình sau:
a,\(\left(x+1\right)^2\ge4x\)
b,\(\left(x+y\right)^2\ge4xy\)
c,\(x+\frac{1}{x}\ge2;a;b>0\)
d,\(\frac{a}{b}+\frac{b}{a}\ge2;a;b>0\)