(\(\frac{6-2\sqrt{2}}{3-\sqrt{2}}-\frac{5}{\sqrt{5}}:\frac{1}{2+\sqrt{5}}\))
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
a)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2(\sqrt{6}+2+\sqrt{6}-2)}{(\sqrt{6}-2)(\sqrt{6}+2)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
b)
\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)
\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
c)
\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left[\frac{\sqrt{2}(\sqrt{3}-1)}{1-\sqrt{3}}-\sqrt{5}\right].(\sqrt{5}-\sqrt{2})\)
\(=(-\sqrt{2}-\sqrt{5})(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)
\(=-(5-2)=-3\)
d)
\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)
9 tháng 8 2019
1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
=\(\frac{\left(\sqrt{15}-\sqrt{5}\right)\cdot\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}+\frac{\left(5-2\sqrt{5}\right)\cdot\left(2\sqrt{5}+4\right)}{\left(2\sqrt{5}-4\right)\cdot\left(2\sqrt{5}+4\right)}\)
=\(\frac{2\sqrt{5}}{2}+\frac{2\sqrt{5}}{4}\)
=\(\sqrt{5}+\frac{\sqrt{5}}{2}\)
=\(\frac{3\sqrt{5}}{2}\)
2/\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
=\(\frac{\left(\sqrt{15}-\sqrt{12}\right)\cdot\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\cdot\left(\sqrt{5}+2\right)}+\frac{\left(6+2\sqrt{6}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+2\right)\cdot\left(\sqrt{3}-2\right)}\)
=\(\frac{\sqrt{3}}{1}+\frac{2\sqrt{3}}{1}\)
=\(3\sqrt{3}\)
3/\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
=\(\frac{\sqrt{3}\cdot\left(3+2\sqrt{3}\right)}{3}+\frac{\left(2+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\cdot\left(\sqrt{2}-1\right)}-\left(2+\sqrt{3}\right)\)
=\(\frac{6+3\sqrt{3}}{3}+\sqrt{2}-\left(2-\sqrt{3}\right)\)
=\(\frac{3\cdot\left(2+\sqrt{3}\right)}{3}+\sqrt{2}-\left(2+\sqrt{3}\right)\)
=\(2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)
=\(\sqrt{2}\)
Câu số 4 bạn có chắc là đúng đề bài không ạ ? Xem lại đề giúp mình nhé, cảm ơn bạn ^^
TN
15 tháng 6 2017
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
S
3 tháng 7 2019
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
S
3 tháng 7 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)
27 tháng 8 2020
a) Ta có: \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)
\(=\frac{\sqrt{5}-2}{\sqrt{5}\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}+\frac{\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}-2-\sqrt{5}+\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{1}{\sqrt{5}+2}\)
b) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{2}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}-\frac{2\sqrt{2}\cdot\left(\sqrt{3}+2\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{3\sqrt{2}+\sqrt{6}+\sqrt{2}\cdot\left(5+3\sqrt{3}\right)-2\sqrt{6}-4\sqrt{2}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{-\sqrt{2}-\sqrt{6}+5\sqrt{2}+3\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{4\sqrt{2}+2\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{2\sqrt{2}\cdot\left(2+\sqrt{3}\right)}{\sqrt{3}\cdot\sqrt{2}\cdot\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{2}{3+\sqrt{3}}\)
21 tháng 7 2017
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
a,ĐK:x≥−12PT⇔√3x+4=√2x+1+1⇔3x+4=2x+2+2√2x+1⇔x+2=2√2x+1⇔x2+4x+4=8x+4⇔x2−4x=0⇔[x=0(tm)x=4(tm)b,ĐK:x≥1PT⇔√2x−1=2√x−1−1⇔2x−1=4x−3−4√x−1⇔2x−2−4√x−1=0⇔x−1−2√x−1=0⇔√x−1(√x−1−2)=0⇔[x−1=0x−1=4⇔[x=1(tm)x=5(tm)