ĐKXĐ:

\(x-1\ge0\)và \(x-2\sqrt{x-1}\ge0\)

<=>\(x\ge1\)và\(x\ge2\sqrt{x-1}\)

<=>\(x\ge1\)và \(x^2\ge4x-4\)

<=>\(x\ge1\)và \(\left(x-4\right)^2\ge0\)( luôn đúng với mọi x)

<=> \(x\ge1\)

\(P=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)

\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)

nếu \(\sqrt{x-1}-1\le0\)thì

\(P=1-\sqrt{x-1}+\sqrt{x-1}+1=2\)

nếu \(\sqrt{x-1}-1\ge0\)thì:

\(P=\sqrt{x-1}-1+\sqrt{x-1}+1=2\sqrt{x-1}\)

∑ = tổng hoán vi

\(Gt\Rightarrow\text{ ∑}\frac{1}{xy}=1\).Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\) ta có:

\(ab+bc+ca=1\)

\(\text{∑}\frac{1}{\sqrt{1+x^2}}=\text{∑}\frac{a}{\sqrt{1+a^2}}=\text{∑}\frac{a}{\sqrt{ab+bc+ca+a^2}}\)

\(=\text{∑}\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\text{∑}\left(\frac{a}{a+b}+\frac{a}{c+a}\right)=\frac{3}{2}\)

đề thế này à .Cho x,y,z>0 thỏa mãn x+y+z=xyz.

Chứng minh \(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\le\frac{3}{2}\)

mấy bài này max dễ bn đăng từng phần 1 mk lm cho

về việc đổi |x-1|² thành (x-1)² khá đơn giản

|x-1|>=0 =>|x-1|^2>=0

(x-1)^2>=0

a)x + (x + 1) + (x + 2) + (x + 3) + ... + (x + 30) = 1240

\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+30\right)=1240\)

\(\Leftrightarrow31x+465=1240\)

\(\Leftrightarrow31x=775\Leftrightarrow x=25\)

b)2|x - 1|² - 3 = 5

\(\Leftrightarrow2\left(x-1\right)^2=8\)

\(\Leftrightarrow\left(x-1\right)^2=4\)

\(\Leftrightarrow\left(x-1\right)^2=2^2=\left(-2\right)^2\)

\(\Leftrightarrow x-1=\pm2\)

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