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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 => \(2-\sqrt{x}\inƯ\left(2\right)\Rightarrow2-\sqrt{x}\in\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\Leftrightarrow x\in\left\{0;1;9;16\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ge5\)
\(-\sqrt{x}\le0\) với mọi \(x\ge5\)
\(\sqrt{x-5}+\sqrt{x+7}>0\)với mọi \(x\ge5\)
Vậy PT trên không tồn tại nghiệm số thực
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}\)
\(=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=c+a\\2c=a+b\end{cases}}\Rightarrow3a=3b=3c=a+b+c\Rightarrow a=b=c\)
\(\Rightarrow BT=\frac{\left(2a\right)^3}{a^3}+\frac{\left(2b\right)^3}{b^3}+\frac{\left(2c\right)^3}{c^3}=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-2}\left(x\ge0,x\ne4\right)\)
\(A=\frac{\sqrt{x}-2+4}{\sqrt{x}-2}=1+\frac{4}{\sqrt{x}-2}\)
+ 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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{2x-\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{3x\sqrt{x}-2x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(\frac{\sqrt{x}^3-x+\sqrt{x}-x+\sqrt{x}-1+2\sqrt{x}^3-x-\sqrt{x}+2x-\sqrt{x}-1-3\sqrt{x}^3+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(\frac{1}{\sqrt{x}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(11:\)
\(\frac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\left(B\right)\)
\(12:B\)
\(13:\sqrt{25x}-\sqrt{9x}=8\)
\(\sqrt{25}\sqrt{x}-\sqrt{9}\sqrt{x}=8\)
\(\sqrt{x}\left(5-3\right)=8\)
\(\sqrt{x}=4< =>x=16\left(C\right)\)
\(14:\frac{4}{\sqrt{5}-1}-\sqrt{5}\)
\(\frac{4-5+\sqrt{5}}{\sqrt{5}-1}\)
\(\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\left(B\right)\)
\(15:\)
\(-\sqrt{a^2\frac{b}{a}}\)
\(-\sqrt{a.b}\left(C\right)\)
\(1:4\left(B\right)\)
\(16:\sqrt{12}-\sqrt{27}+\sqrt{3}\)
\(\sqrt{3}\left(\sqrt{4}-\sqrt{9}+1\right)\)
\(\sqrt{3}\left(2-3+1\right)=0\left(B\right)\)
\(17:\sqrt{18}+\frac{2}{\sqrt{2}}-3\sqrt{8}\)
\(\sqrt{2}\left(\sqrt{9}+1-3\sqrt{4}\right)\)
\(\sqrt{2}.2=2\sqrt{2}\left(D\right)\)
\(18:\sqrt{x^2}=\left|x\right|=13\)
\(x=\pm13\left(D\right)\)
\(19:\left|x-1\right|\left(C\right)\)
\(20:\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\left|\sqrt{5}-2\right|=\sqrt{5}-2\left(B\right)\)
hok tốt
xét tam giác ABC vuông tại A có:
AB2+AC2=BC2 (định lí pytago)
hay 32+72=BC2
=>BC=\(\sqrt{3^2+7^2}=\sqrt{58}\)
ta lại có: x.BC=AB.AC
hay x=3.7:\(\sqrt{58}\)=\(\frac{21}{\sqrt{58}}\)
vậy x=...
y=...
Xét tam giác ABC vuông tại A, đường cao AH ( tạm gọi chân đường cao là H nhé )
Áp dụng định lí Pytago cho tam giác ABC vuông tại A
\(BC^2=AB^2+AC^2=9+49=58\Rightarrow BC=\sqrt{58}\)cm
hay \(y=\sqrt{58}\)cm
* Áp dụng hệ thức : \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{9}+\frac{1}{49}=\frac{58}{441}\)
\(\Rightarrow AH^2=\frac{441}{58}\Leftrightarrow AH=\frac{21\sqrt{58}}{58}\)cm hay \(x=\frac{21\sqrt{58}}{58}\)cm