\(=\left(-1\right)\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}\)

\(=\left(-1\right)\cdot\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)\)

\(=\left(-\sqrt{3}-1\right)+\left(\sqrt{3}-1\right)\)

\(=-2\)

CM gium minh

CHo thêm a,b,c dương nữa nhé

Áp dụng BĐT AM-GM ta có:

\(\frac{a^2+c^2}{b}\ge\frac{2ca}{b};\frac{b^2+a^2}{c}\ge\frac{2ab}{c};\frac{b^2+c^2}{a}\ge\frac{2bc}{a}\)

Cần cm \(\frac{2ab}{c}+\frac{2bc}{a}+\frac{2ca}{b}\ge2\left(a+b+c\right)\)

Hay \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)

Áp dụng tiếp AM-GM có:

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)

Tương tự ta cũng có:

\(\frac{bc}{a}+\frac{ca}{b}\ge2c;\frac{ab}{c}+\frac{ca}{b}\ge2a\)

Cộng theo vế 3 BĐT trên rồi thu gọn ta có:

 \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\) ( đúng)

Hay ta có ĐPCM

Tiếp =))

c)Áp dụng BĐT AM-GM ta có:

\(x\sqrt{y-1}\le\frac{x\left(y-1+1\right)}{2}=\frac{xy}{2}\)

\(2y\sqrt{x-1}\le\frac{2y\left(x-1+1\right)}{2}=\frac{2xy}{2}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=x\sqrt{y-1}+2y\sqrt{x-1}\le\frac{3xy}{2}=VP\)

Nên xảy ra khi \(x=y\) thay vào giải ra có: x=y=2

d)\(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)

\(pt\Leftrightarrow\sqrt{2x^2+x+1}-2+\sqrt{x^2-x+1}-1=3x-3\)

\(\Leftrightarrow\frac{2x^2+x+1-4}{\sqrt{2x^2+x+1}+2}+\frac{x^2-x+1-1}{\sqrt{x^2-x+1}+1}=3\left(x-1\right)\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x\left(x-1\right)}{\sqrt{x^2-x+1}+1}-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{\left(2x+3\right)}{\sqrt{2x^2+x+1}+2}+\frac{x}{\sqrt{x^2-x+1}+1}-3\right)=0\)

pt trong ngoặc vn nên x=1

Tắm đã làm nốt cho :))

Chả ai giúp t gank =)), mà lần sau đăng ít 1 thôi đăng lắm thế này nhìn nản cmn luôn ấy

a)\(\sqrt{x^2+x-5}+\sqrt{-x^2+x+3}=x^2-3x+4\)

\(pt\Leftrightarrow\sqrt{x^2+x-5}-1+\sqrt{-x^2+x+3}-1=x^2-3x+2\)

\(\Leftrightarrow\frac{x^2+x-5-1}{\sqrt{x^2+x-5}+1}+\frac{-x^2+x+3-1}{\sqrt{-x^2+x+3}+1}=\left(x-1\right)\left(x-2\right)\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x+3\right)}{\sqrt{x^2+x-5}+1}+\frac{-\left(x-2\right)\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{\left(x+3\right)}{\sqrt{x^2+x-5}+1}-\frac{\left(x+1\right)}{\sqrt{-x^2+x+3}+1}-\left(x-1\right)\right]=0\)

Pt trong ngoặc <0 nên x=2 là nghiệm

b)\(\frac{x^2}{2}+\frac{x}{2}+1=\sqrt{2x^3-x^2+x+1}\)\

Đk:\(x\ge-\frac{1}{2}\)

\(\Leftrightarrow\frac{x^2}{2}+\frac{x}{2}+1-\left(2x+1\right)=\sqrt{2x^3-x^2+x+1}-\left(2x+1\right)\)

\(\Leftrightarrow\frac{x^2}{2}+\frac{x}{2}+1-\left(2x+1\right)=\frac{2x^3-x^2+x+1-\left(2x+1\right)^2}{\sqrt{2x^3-x^2+x+1}+2x+1}\)

\(\Leftrightarrow\frac{x^2-3x}{2}-\frac{2x^3-5x^2-3x}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)

\(\Leftrightarrow\frac{x\left(x-3\right)}{2}-\frac{x\left(x-3\right)\left(2x+1\right)}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)

\(\Leftrightarrow x\left(x-3\right)\left(\frac{1}{2}-\frac{2x+1}{\sqrt{2x^3-x^2+x+1}+2x+1}\right)=0\)

Pt trong ngoặc vô nghiệm nốt nên 

\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Gọi cái vế trái của BĐT cần c/m là P

Áp dụng  BĐT Cô-si dạng  \(\frac{1}{a+b+c+x+y+z}\le\frac{1}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Đẳng thức xảy ra  \(\Leftrightarrow\)  a = b = c = x = y = z

và  \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

Đẳng thức xảy ra  \(\Leftrightarrow\)  a = b = c = x = y = z

Ta có  \(\frac{1}{10a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+\left(a+a\right)+\left(a+a\right)+\left(a+a\right)+\left(a+a\right)}\)

\(\le\frac{1}{36}\left(\frac{1}{a+b}+\frac{1}{a+c}+4.\frac{1}{a+a}\right)\le\frac{1}{36}\left[\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{c}\right)+\frac{2}{a}\right]\)

\(=\frac{1}{36}\left[\frac{1}{4}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{2}{a}\right]\)   (1)

Tương tự  \(\frac{1}{10b+c+a}\le\frac{1}{36}\left[\frac{1}{4}\left(\frac{2}{b}+\frac{1}{c}+\frac{1}{a}\right)+\frac{2}{b}\right]\)   (2)

và   \(\frac{1}{10c+a+b}\le\frac{1}{36}\left[\frac{1}{4}\left(\frac{2}{c}+\frac{1}{a}+\frac{1}{b}\right)+\frac{2}{c}\right]\)   (3)

Cộng (1), (2), (3) vế theo vế ta được

\(P\le\frac{1}{36}\left[\frac{1}{4}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)+\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)\right]=...=\frac{1}{12}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Kết hợp  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{6}\)  (theo đề bài) và BĐT  \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)

Ta có  \(P^2\le\frac{1}{144}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{144}\left[\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\right]\)

\(\le\frac{1}{144}\left(\frac{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{6}+\frac{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\right)\)

Suy ra  \(P^2\le\frac{1}{144}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{1}{144}\left(\frac{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{6}+\frac{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\right)\)

Đặt  \(t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)  thì  \(\frac{1}{144}t^2\le\frac{1}{144}\left(\frac{1+t}{6}+\frac{2t^2}{3}\right)\)

\(\Leftrightarrow\)  \(2t^2-t-1\le0\)  \(\Leftrightarrow\)  \(\frac{-1}{2}\le t\le1\)

Do đó  \(P^2\le\frac{1}{144}t^2\le\frac{1}{144}.1^2=\frac{1}{144}\)  \(\Rightarrow\)  \(P\le\frac{1}{12}\)

Đẳng thức xảy ra  \(\Leftrightarrow\)  \(a=b=c=3\)

mk nhầm cái đoạn  \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)  đẳng thức xảy ra  \(\Leftrightarrow\)  a = b