kết bạn nhé

mình con trai cưa có gấu đẹp trai sáu muối con nhà giàu

Haha ok bn :)

\(\text{Ta có : }(a-b)^2\ge0\)

\(\Leftrightarrow a^2+b^2-2ab\ge0(\text{*})\)

\(\text{Ta có :}2a(\sqrt{b}-\frac{1}{2})^2\ge0\text{ do a là số thực dương}\)

\(\Rightarrow2a(b-\sqrt{b}+\frac{1}{4})\ge0\)

\(\Leftrightarrow2ab-2a\sqrt{b}+\frac{a}{2}\ge0\text{(**)}\)

\(\text{Ta có : }2b(\sqrt{a}-\frac{1}{2})^2\ge0\text{ do b là số thực dương }\)

\(\Leftrightarrow2b(a-\sqrt{a}+\frac{1}{4})\ge0\)

\(\Leftrightarrow2ab-ab\sqrt{a}+\frac{b}{2}\ge0(\text{***})\)

Cộng (*), (**) và (***) vế theo vế, ta có:

\(a^2+b^2-2ab+2ab-2a\sqrt{b}+\frac{a}{2}+2ab-2b\sqrt{a}+\frac{b}{2}\ge0\)

\(a^2+b^2+2ab+\frac{a+b}{2}-(2a\sqrt{b}+2b\sqrt{a})\ge0\)

\(\Rightarrow(a+b)^2+\frac{a+b}{2}\ge2a\sqrt{b}+2b\sqrt{a}(đpcm)\)

Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)

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

b) Để \(A< -1\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)

\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

Ta có: \(a< a+b\left(a,b>0\right)\Rightarrow\frac{a}{a+b}< 1\)

Có: \(\frac{a}{a+b}=\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}\)

Lại có: \(\frac{a}{b+a}< 1\Leftrightarrow\sqrt{\frac{a}{b+a}}< 1\Rightarrow\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}< \sqrt{\frac{a}{a+b}}\Rightarrow\frac{a}{a+b}< \sqrt{\frac{a}{a+b}}\)

Chứng minh tương tự ta có:

\(\frac{b}{b+c}< \sqrt{\frac{b}{b+c}}\)

\(\frac{c}{c+a}< \sqrt{\frac{c}{c+a}}\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\)

                                                                                               đpcm

Sai thì thôi nhé~

Mới lp 8