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câu b trc nha
B = \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{2}.\sqrt{3}+2\sqrt{2}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{2+2+\sqrt{2}+2\sqrt{2}-\sqrt{3}-\sqrt{6}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)+2\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}+1\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\left(\sqrt{2}+1\right)\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\sqrt{2}\) + 1
A = \(\dfrac{21}{2}\) . (\(\sqrt{4+2\sqrt{3}}\) + \(\sqrt{6-2\sqrt{5}}\) )2 - 15\(\sqrt{15}\)
- 3(\(\sqrt{4-2\sqrt{3}}\) +\(\sqrt{6+2\sqrt{5}}\) )2
= \(\dfrac{21}{2}\).(\(\sqrt{\left(\sqrt{3}+1\right)^2}\) + \(\sqrt{\left(\sqrt{5}-1\right)^2}\))2-15\(\sqrt{15}\)
-3(\(\sqrt{\left(\sqrt{3}-1\right)^2}\) + \(\sqrt{\left(\sqrt{5}+1\right)^2}\))2
= \(\dfrac{21}{2}\).(\(\sqrt{3}\) +1+ \(\sqrt{5}\) - 1)2 -3.(\(\sqrt{3}\) - 1 + \(\sqrt{5}\) +1)2
- 15\(\sqrt{15}\)
= \(\dfrac{21}{2}\).(8+2\(\sqrt{15}\) ) - 3(8 + 2\(\sqrt{15}\) ) -15\(\sqrt{15}\)
= \(\dfrac{15}{2}\) .2.(4+\(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.( 4 + \(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.(4+\(\sqrt{15}\) -\(\sqrt{15}\)) =15.4 = 60
Vậy A = 60.
\(=\left(5\sqrt{2}-\sqrt{5}\right)\left(3\sqrt{2}-2\sqrt{5}+2\sqrt{2}\right)\)
\(=\left(5\sqrt{2}-\sqrt{5}\right)\left(5\sqrt{2}-2\sqrt{5}\right)\)
\(=60-15\sqrt{10}\)
Bạn tham khảo lời giải tại đây:
Câu hỏi của Toán Chuyên Học - Toán lớp 9 | Học trực tuyến
\(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
P>3/2
=>P-3/2>0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
=>-căn x+2>0
=>-căn x>-2
=>0<x<4
ban co the nao giai chi tiet cho minh dc ko
a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{-x+9}\)
\(=\dfrac{5\left(\sqrt{x}-3\right)}{x-9}=\dfrac{5}{\sqrt{x}+3}\)
b: Để A<1 thì A-1<0
\(\Leftrightarrow5-\sqrt{x}-3< 0\)
=>2-căn x<0
=>căn x>2
=>x>4
\(taco:A^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}}\Rightarrow A^2-A=2\Rightarrow A=2\left(dpcm\right)\)
\(A=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}}}\)
\(\Rightarrow A^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}}\)
\(\Rightarrow A^2-A=\left(2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}\right)-\left(\sqrt{2+\sqrt{2+\cdot\cdot\cdot\cdot}}\right)\)
\(\Rightarrow A=2\)