a,  \(14\sqrt{\frac{1}{7}}-\frac{3}{2}\sqrt{28}\)\(+20\sqrt{0,63}\)

\(=2\sqrt{7}-3\sqrt{7}+6\sqrt{7}\)

\(=5\sqrt{7}\)

b,  \(\sqrt{\frac{a}{2}}+\frac{4}{5}\sqrt{8\text{a}}-\sqrt{\frac{2\text{a}}{9}}v\text{ới}a\ge0\)

\(=\sqrt{\frac{2\text{a}}{4\text{ }}}+\frac{8}{5}\sqrt{2\text{a}}-\frac{1}{3}\sqrt{2\text{a}}\)

\(=\frac{1}{2}\sqrt{2\text{a}}+\frac{8}{5}\sqrt{2\text{a}}-\frac{1}{3}\sqrt{2\text{a}}\)

\(=\frac{53}{30}\sqrt{2\text{a}}\)

Bạn kiểm tra lại điều kiện của đề bài nhé vì x ở dưới mẫu ko thể = 0 được

a) M = \(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\): \(\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\)  ( ĐKXĐ : x > 0 , x\(\ne\)1)

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

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

M = \(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

M = \(\frac{x}{\sqrt{x}-1}\)

b) M = \(\frac{x}{\sqrt{x}-1}\)( ĐKXĐ : x > 0, x\(\ne\)1)

    Ta có : \(\frac{x}{\sqrt{x}-1}\)= \(\frac{-1}{2}\)

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

\(\Leftrightarrow\) 2x + \(\sqrt{x}\)-1 = 0

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=-1\end{cases}\Rightarrow}x=\frac{1}{4}\left(TM\text{Đ}K\right)\)

Vậy x =\(\frac{1}{4}\) để M = \(\frac{-1}{2}\)

ĐKXĐ : \(\hept{\begin{cases}x-1\ge0\\2\text{x}-1\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge1\\x\ge\frac{1}{2}\end{cases}}}\)\(\Rightarrow\)\(x\ge\frac{1}{2}\)( do \(\frac{1}{2}< 1\) nên ta đặt những giá trị lớn hơn
\(\frac{1}{2}\)sẽ bao hàm cả những giá trị lớn hơn 1 luôn nhé )

Đây là mình giải thích kĩ cho bạn nhé ko thì bạn chỉ cần ghi đkxđ là \(x\ge\frac{1}{2}\) thôi

Bài 2 : 

a, ĐK : \(x\ne-3\)

b, \(B=\left(\frac{x+2}{x+3}-\frac{x^2+4x}{x^2+6x+9}\right):\left(\frac{3}{x+3}+1\right)\)

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

\(=\frac{x^2+5x+6-x^2-4x}{\left(x+3\right)}:\left(\frac{x+6}{x+3}\right)=\frac{x+6}{x+3}.\frac{x+3}{x+6}=1\)

Bài 3 : 

a, Với \(x\ne\pm3\)

\(C=\left(\frac{x^2}{x^3+27}-\frac{1}{x+3}\right):\left(\frac{-2x+6+x^2}{x^2-3x+9}-1\right)\)

\(=\left(\frac{x^2}{\left(x+3\right)\left(x^2-3x+9\right)}-\frac{1}{x+3}\right):\left(\frac{x^2-2x+6-x^2+3x-9}{x^2-3x+9}\right)\)

\(=\left(\frac{x^2-x^2+3x-9}{\left(x+3\right)\left(x^2-3x+9\right)}\right):\left(\frac{x-3}{x^2-3x+9}\right)=\frac{3\left(x-3\right)\left(x^2-3x+9\right)}{\left(x+3\right)\left(x-3\right)\left(x^2-3x+9\right)}=\frac{3}{x+3}\)

b, Ta có : \(C>\frac{1}{5}\Rightarrow\frac{3}{x+5}-\frac{1}{5}>0\Leftrightarrow\frac{10-x}{5\left(x+5\right)}>0\)

TH1 : \(\hept{\begin{cases}10-x>0\\5\left(x+5\right)>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 10\\x+5>0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< 10\\x>-5\end{cases}\Leftrightarrow}-5< x< 10}\)

TH2 : \(\hept{\begin{cases}10-x< 0\\5\left(x+5\right)< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>10\\x+5< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>10\\x< -5\end{cases}}}\)vô lí 

1234583 +  987736489 = 988971072

988971072 bạn nhé

a, \(P=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

\(P=\frac{x-1}{\sqrt{x}}:\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

b, \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) (thỏa mãn)

thay vào P ta đc : 

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

\(P=\frac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}\) vì \(\sqrt{3}-1>0\)

\(P=\frac{\sqrt{3}^2}{\sqrt{3}-1}=\frac{3}{\sqrt{3}-1}\)

c, \(P-2=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-2=\frac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\frac{x+1}{\sqrt{x}}\) mà x > 0

\(\Rightarrow P-2>0\Leftrightarrow P>2\)

d, thay P vào pt ta đc : \(\left(\sqrt{x}+1\right)^2=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x-4\sqrt{x}+4+\sqrt{x-4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

mà \(\hept{\begin{cases}\left(\sqrt{x}-2\right)^2\ge0\\\sqrt{x-4}\ge0\end{cases}}\) nên \(\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow x=4\left(tm\right)\)

vậy_

đội tuyển toán à mà toàn đăng bài khó thế :((

ukm khổ ghe

theo cô si có \(ab\le\frac{a^2+b^2}{2}\Rightarrow a^2+ab+b^2\le a^2+\frac{a^2+b^2}{2}+b^2\)

\(\Rightarrow\frac{a^5}{a^2+ab+b^2}\ge\frac{a^5}{\frac{3}{2}\left(a^2+b^2\right)}=\frac{2}{3}\cdot\frac{a^5}{a^2+b^2}\)

tương tự cm đc \(\frac{b^5}{b^2+bc+c^2}\ge\frac{2}{3}\cdot\frac{b^5}{b^2+c^2}\) và \(\frac{c^5}{c^2+ac+a^2}\ge\frac{2}{3}\cdot\frac{c^5}{c^2+a^2}\)

\(\Rightarrow VT\ge\frac{2}{3}\left(\frac{a^5}{a^2+b^2}+\frac{b^5}{b^2+c^2}+\frac{c^5}{c^2+a^2}\right)\)

\(\Rightarrow VT\ge\frac{2}{3}\left(\frac{a^3\left(a^2+b^2\right)-a^3b^2}{a^2+b^2}+\frac{b^3\left(b^2+c^2\right)-b^3c^2}{b^2+c^2}+\frac{c^3\left(c^2+a^2\right)-c^3a^2}{c^2+a^2}\right)\)

áp dụng cô si ta có \(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{cases}\Rightarrow\hept{\begin{cases}\frac{-a^3b^2}{a^2+b^2}\ge\frac{a^2b}{2}\\\frac{-b^3c^2}{b^2+c^2}\ge\frac{b^2c}{2}\\\frac{-c^3a^2}{c^2+a^2}\ge\frac{c^2a}{2}\end{cases}}}\)

\(\Rightarrow VT\ge\frac{2}{3}\left(a^3+\frac{a^2b}{2}+b^3+\frac{b^2c}{2}+c^3+\frac{c^2a}{2}\right)\)

hjhj khó quá :v

ủa hình như chưa làm đc đúng k