Rút gọn biểu thức :
\(A = {12\over √3+√2+√5}\)
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Ta có: \(\left(\sqrt{12}-2\sqrt{18}+5\sqrt{3}\right)\cdot\sqrt{3}+5\sqrt{6}\)
\(=\left(2\sqrt{3}-6\sqrt{3}+5\sqrt{3}\right)\cdot\sqrt{3}+5\sqrt{6}\)
\(=3+5\sqrt{6}\)
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
a ⇒A=\(4\sqrt{4\times3}+3\sqrt{25\times3}-5\sqrt{16\times3}=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}=3\sqrt{3}\)
b ĐKXĐ x≥2 ⇔\(\sqrt{x-2}+3\sqrt{x-2}=16\Leftrightarrow4\sqrt{x-2}=16\Leftrightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Leftrightarrow x=18\)
a. \(A=4\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)
\(=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)
\(=3\sqrt{3}\)
b. \(\sqrt{x-2}-\sqrt{9x-18}=16\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{9\left(x-2\right)}=16\)
\(\Leftrightarrow\sqrt{x-2}-3\sqrt{x-2}=16\)
\(\Leftrightarrow-2\sqrt{x-2}=16\)
\(\Leftrightarrow\sqrt{x-2}=-8\) ( Vô lý )
Vậy PT vô nghiệm
\(x=\frac{12}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
=> \(x^2=\left(\frac{12}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\right)^2\)
<=> \(x^2=\frac{144}{3+2+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
<=> \(x^2=\frac{144}{2\left(5+\sqrt{6}+\sqrt{10}+\sqrt{15}\right)}\)
<=> \(x^2=\frac{144}{2\left[\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)+\sqrt{3}\left(\sqrt{5}+\sqrt{2}\right)\right]}\)
<=> \(x^2=\frac{144}{2\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
<=> \(x^2=\frac{72}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
=> \(x=\frac{6\sqrt{2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}+\sqrt{3}\right)}}\)
Cho biểu thức
A= (\( {1 \over x-2}\)+\({1 \over x+2}\)) : \( {5-x \over x-2}\)
a) Tìm ĐKXĐ
b) Rút gọn A
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
Mình viết nhầm bài là:
Rút gọn biểu thức: \(A = {{12} \over√3+√2+√5}\)