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cot B = \(\dfrac{5}{13}=>tanB=\dfrac{13}{5}\)
AC=AB.tanB
AC= 15.\(\dfrac{13}{5}\)
AC= 39cm
BC2=AB2+AC2
BC2=225+1521=1746
BC=3 \(\sqrt{194}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\left(\sqrt{a}+2\right)\)
=0
b) Ta có: \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{a-6\sqrt{a}+9}{\sqrt{a}-3}\)
\(=3-\sqrt{a}-\sqrt{a}+3\)
\(=6-2\sqrt{a}\)
c) Ta có: \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)
=0
d) Ta có: \(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\)
\(=a+\sqrt{ab}+b+\sqrt{ab}\)
\(=a+2\sqrt{ab}+b\)
Bài 1:
a.
\(\frac{4\sqrt{5}+\sqrt{15}}{\sqrt{5}}=\frac{\sqrt{5}(4+\sqrt{3})}{\sqrt{5}}=4+\sqrt{3}\)
$\frac{7-\sqrt{7}}{3\sqrt{7}}=\frac{\sqrt{7}(\sqrt{7}-1)}{3\sqrt{7}}=\frac{\sqrt{7}-1}{3}$
\(\frac{4\sqrt{2}-\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}(4-\sqrt{3})}{\sqrt{2}.\sqrt{6}}=\frac{4-\sqrt{3}}{\sqrt{6}}\)
\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\frac{(3\sqrt{2}-2\sqrt{3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{\sqrt{6}}{3-2}=\sqrt{6}\)
b.
\(\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}(\sqrt{a}-2)}{\sqrt{a}-2}=\sqrt{a}\)
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}=\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
\(\frac{a+10\sqrt{a}+25}{\sqrt{a}+5}=\frac{(\sqrt{a}+5)^2}{\sqrt{a}+5}=\sqrt{a}+5\)
\(\frac{a-9}{\sqrt{a}+3}=\frac{(\sqrt{a}-3)(\sqrt{a}+3)}{\sqrt{a}-3}=\sqrt{a}+3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
b: Ta có: \(2\sqrt{3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{18}\)
mà 12<18
nên \(2\sqrt{3}< 3\sqrt{2}\)
c: Ta có: \(\dfrac{12}{\sqrt{5}-2}=12\sqrt{5}+24\)
d: Ta có: \(\dfrac{24}{2-x}\cdot\sqrt{\dfrac{x^2-4x+4}{36}}\)
\(=\dfrac{24}{2-x}\cdot\dfrac{2-x}{6}\)
=4
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: Ta có: ΔABC vuông tại A
nên \(\widehat{B}+\widehat{C}=90^0\)
hay \(\widehat{B}=60^0\)
Xét ΔABC vuông tại A có
\(AB=AC\cdot\tan30^0\)
\(=\dfrac{10\sqrt{3}}{3}\left(cm\right)\)
Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:
\(AB^2+AC^2=BC^2\)
\(\Leftrightarrow BC=\dfrac{20\sqrt{3}}{3}\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Thay x=9 vào A, ta được:
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)
b) Ta có: M=B:A
\(=\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}-5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(=\dfrac{x+3\sqrt{x}+\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\dfrac{x+4\sqrt{x}+5}{x+7\sqrt{x}+10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn ơi, làm như vậy thì quá ngắn rồi ạ, với lại bạn làm thiếu mất đề bài của mình rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{15+5\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{5}\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{6-2\sqrt{5}}{2}}\)
\(=\sqrt{5}\sqrt{\dfrac{6+2\sqrt{5}}{2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}=\sqrt{5}\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{2}}-\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{2}}\)
\(=\sqrt{5}.\dfrac{\left|\sqrt{5}+1\right|}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{5}.\dfrac{\sqrt{5}+1}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}}\)
\(=\dfrac{5+\sqrt{5}-\sqrt{5}+1}{\sqrt{2}}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}=\dfrac{5\sqrt{3}}{\sqrt{15}}+\dfrac{3\sqrt{5}}{\sqrt{15}}=\dfrac{5\sqrt{3}}{\sqrt{5}.\sqrt{3}}+\dfrac{3\sqrt{5}}{\sqrt{3}.\sqrt{5}}=\sqrt{5}+\sqrt{3}\)
\(h=2R\)
\(V=h.\pi R^2=2R.\pi R^2=16\pi\)
\(\Rightarrow R^3=8\Rightarrow R=2\Rightarrow h=4\)
\(S_{tp}=2\pi R^2+2\pi Rh=24\pi\) \(\left(cm^2\right)\)