a) \(B=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\div\frac{\sqrt{x}+2}{x-4}\)

\(=\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{x-4}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}+2}{x-4}\cdot\frac{x-4}{\sqrt{x}+2}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)

b) \(C=A\left(B-2\right)=\frac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\frac{-2}{\sqrt{x}+2}=\frac{2}{2-\sqrt{x}}\)

Để C nguyên => \(2-\sqrt{x}\inƯ\left(2\right)\Rightarrow2-\sqrt{x}\in\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\Leftrightarrow x\in\left\{0;1;9;16\right\}\)

\(ĐK:x\ge5\)

\(-\sqrt{x}\le0\) với mọi \(x\ge5\)

\(\sqrt{x-5}+\sqrt{x+7}>0\)với mọi \(x\ge5\)

Vậy PT trên không tồn tại nghiệm số thực

\(\sqrt{9\left(a-5\right)^2}=3\left|a-5\right|\)  

Với \(a\ge5;\sqrt{9\left(a-5\right)^2}=3a-15\)

Với 0 < a < 5 ; \(\sqrt{9\left(a-5\right)^2}=15-3a\)

Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}\)

\(=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=c+a\\2c=a+b\end{cases}}\Rightarrow3a=3b=3c=a+b+c\Rightarrow a=b=c\)

\(\Rightarrow BT=\frac{\left(2a\right)^3}{a^3}+\frac{\left(2b\right)^3}{b^3}+\frac{\left(2c\right)^3}{c^3}=24\)

\(A=\frac{\sqrt{x}+2}{\sqrt{x}-2}\left(x\ge0,x\ne4\right)\)

\(A=\frac{\sqrt{x}-2+4}{\sqrt{x}-2}=1+\frac{4}{\sqrt{x}-2}\)

+ Nếu x ko là SCP

=> \(\sqrt{x}\notin Z\Rightarrow\frac{4}{\sqrt{x}-2}\notin Z\) (loại)

+ Nếu x là SCP

\(\Rightarrow\sqrt{x}-2\in Z\)

Để A nguyên thì \(\frac{4}{\sqrt{x}-2}\in Z\)

Hay \(\sqrt{x}-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Bạn tự lm tiếp nha

người ta giải giúp bạn sao ko k và cảm mơn đi

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

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

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

\(\frac{1}{\sqrt{x}+1}\)

a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

- \(cosx=\frac{1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

- \(cosx=\frac{-1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

\(11:\)

\(\frac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\left(B\right)\)

\(12:B\)

\(13:\sqrt{25x}-\sqrt{9x}=8\)

\(\sqrt{25}\sqrt{x}-\sqrt{9}\sqrt{x}=8\)

\(\sqrt{x}\left(5-3\right)=8\)

\(\sqrt{x}=4< =>x=16\left(C\right)\)

\(14:\frac{4}{\sqrt{5}-1}-\sqrt{5}\)

\(\frac{4-5+\sqrt{5}}{\sqrt{5}-1}\)

\(\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\left(B\right)\)

\(15:\)

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

\(-\sqrt{a.b}\left(C\right)\)

\(1:4\left(B\right)\)

\(16:\sqrt{12}-\sqrt{27}+\sqrt{3}\)

\(\sqrt{3}\left(\sqrt{4}-\sqrt{9}+1\right)\)

\(\sqrt{3}\left(2-3+1\right)=0\left(B\right)\)

\(17:\sqrt{18}+\frac{2}{\sqrt{2}}-3\sqrt{8}\)

\(\sqrt{2}\left(\sqrt{9}+1-3\sqrt{4}\right)\)

\(\sqrt{2}.2=2\sqrt{2}\left(D\right)\)

\(18:\sqrt{x^2}=\left|x\right|=13\)

\(x=\pm13\left(D\right)\)

\(19:\left|x-1\right|\left(C\right)\)

\(20:\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(\left|\sqrt{5}-2\right|=\sqrt{5}-2\left(B\right)\)

hok tốt

chịu luôn

ko dc vì

tui hok lớp 4

áp dụng bđt bunhia- cốp xki với bộ số (a,b,c)(1,1,1)

\(\left(a^2+b^2+c^2\right)\left(1+1+1\right)\ge\left(a+b+c\right)^2\)

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(< =>ĐPCM\)

Áp dụng BĐT Cauchy ta có:
\(a^2+b^2\ge2ab\) ; \(b^2+c^2\ge2bc\) ; \(c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)=\left(a+b+c\right)^2\)

Dấu "=" xảy ra khi: a = b = c