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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 \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
5 tháng 8 2020
1) Sửa đề: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\), ta được:
\(A=\frac{-5\cdot\sqrt{\left(\sqrt{2}-1\right)^2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\)
\(=\frac{-5\cdot\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1+3}\)
\(=\frac{-5\sqrt{2}+5+2}{\sqrt{2}+2}\)
\(=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(A=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
2) Ta có: \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\left(x+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}+x+2\sqrt{x}+2+x+x\sqrt{x}-\sqrt{x}-1-\left(2x+2\sqrt{x}+x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2x\sqrt{x}+\sqrt{x}+1-2x-2\sqrt{x}-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: \(x=7-2\sqrt{6}\)
\(=6-2\sqrt{6}\cdot1+1\)
\(=\left(\sqrt{6}-1\right)^2\)
Thay \(x=\left(\sqrt{6}-1\right)^2\) vào biểu thức \(B=\frac{\sqrt{x}}{x+\sqrt{x}+1}\), ta được:
\(B=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{\left(\sqrt{6}-1\right)^2+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(=\frac{\sqrt{6}-1}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
Vậy: Khi \(x=7-2\sqrt{6}\) thì \(B=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
3) Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\frac{\sqrt{x}\left(x-3\sqrt{x}-x-9\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(-3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\cdot2\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(C=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\), ta được:
\(C=\frac{-3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}}{2\cdot\sqrt{\left(2-\sqrt{3}\right)^2}+4}\)
\(=\frac{-3\cdot\left(2-\sqrt{3}\right)}{2\cdot\left(2-\sqrt{3}\right)+4}\)
\(=\frac{-6+3\sqrt{3}}{4-2\sqrt{3}+4}\)
\(=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(C=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
MT
30 tháng 7 2016
1.
đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)
có \(a^2+b^2=4\)
pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)
vì a,b>o nên \(a-b=\sqrt{2}\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Bình phương 2 vế:
\(4-2\sqrt{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
10 tháng 7 2021
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
28 tháng 8 2017
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
VT
9 tháng 10 2019
a)= \(\left(3+\sqrt{5}\right)\left(\sqrt{\left(3-\sqrt{5}\right)^2}\right)\)=\(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
b)= \(\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{\sqrt{2^2.7}}{2}-2\)=\(\frac{2\left(3-\sqrt{7}\right)}{9-7}+\sqrt{7}-2\)=1
c) =\(\frac{3}{3\left(\sqrt{7}-2\right)}-\frac{3}{3\left(\sqrt{7}+2\right)}\)=\(\frac{1}{\sqrt{7}-2}-\frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2-\left(\sqrt{7}-2\right)}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)}\)=\(\frac{4}{7-4}=\frac{4}{3}\)
d) =\(\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)^{ }\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{\left(88-44\sqrt{3}\right)}{25-3}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{22\left(4-2\sqrt{3}\right)}{22}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)\)=3-1 = 2
e) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)= \(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)= \(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)= \(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)
AN
4 tháng 10 2019
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
vl :>>
thiếu = 2 ạ