\(\sqrt{5}\)+\(\sqrt{6}\))2 -
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13 tháng 7 2018
(Căn 5 + căn 6)^2 = 5 + 2*căn 30 + 6 = 11 + 2*căn 30 Căn 120 = căn 4*30 = 2*căn 30 (Căn 5 + căn 6)^2 - căn 120 = 11
10 tháng 9 2018

\(1.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+\sqrt{84}=\left(3\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}=21-2\sqrt{21}+2\sqrt{21}=21\)

\(2.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)

5 tháng 6 2018

\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)

\(=\frac{5}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{3}\right)^2\)

\(=\frac{5}{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{3}\right)^2\)

\(=\frac{5}{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}\right)^2\)

\(=\frac{5}{2}\left(\sqrt{3}\right)^2+\frac{1}{2}\left(\sqrt{5}\right)^2=\frac{15}{2}+\frac{5}{2}=\frac{20}{2}=10\)

NV
17 tháng 6 2019

\(a\sqrt{b}-b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(7\sqrt{7}+3\sqrt{3}=\left(\sqrt{7}+\sqrt{3}\right)\left(7-\sqrt{21}+3\right)=\left(\sqrt{7}+\sqrt{3}\right)\left(10-\sqrt{21}\right)\)

\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)

\(1-a\sqrt{a}=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)\)

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

\(\left(\sqrt{2}+1\right)^2-4\sqrt{2}=\left(\sqrt{2}-1\right)^2\)

\(\left(\sqrt{5}+2\right)^2-8\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

2 cái trên đều áp dụng HĐT \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)

\(5\sqrt{2}-2\sqrt{5}=\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)\)

20 tháng 7 2016

\(\frac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}\)

\(\frac{\sqrt{2^2-2\sqrt{5}2+\sqrt{5^2}}}{2-\sqrt{5}}\)

\(\frac{\sqrt{\left(2-\sqrt{5}\right)^2}}{2-\sqrt{5}}\)

\(\frac{\sqrt{5}-2}{2-\sqrt{5}}\)

= -1

Chúc bạn làm bài tốt :)

3 tháng 8 2016

\(\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=3\)

3 tháng 8 2016

\(\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)

\(=\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}\)

\(=3\)

(Tíck cho mìk vs nhé!)

NV
11 tháng 8 2020

5.

ĐKXĐ: ...

\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)

\(\Leftrightarrow x=5\)

6.

ĐKXĐ: \(-4\le x\le4\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{4-x}+2=2\sqrt{x+4}+4\)

\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)

\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)

\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)

\(\Rightarrow x=\frac{96}{25}\)

NV
11 tháng 8 2020

1.

Bạn coi lại đề

2.

ĐKXĐ: \(1\le x\le2\)

Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:

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

\(\Leftrightarrow3\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)

\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)

\(\Leftrightarrow3\sqrt{2-x}+2-\sqrt{x+2}+1-\sqrt{x-1}=0\)

\(\Leftrightarrow3\sqrt{2-x}+\frac{2-x}{2+\sqrt{x+2}}+\frac{2-x}{1+\sqrt{x-1}}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(3+\frac{\sqrt{2-x}}{2+\sqrt{x+2}}+\frac{\sqrt{2-x}}{1+\sqrt{x-1}}\right)=0\)

\(\Leftrightarrow\sqrt{2-x}=0\Rightarrow x=2\)