HV
12
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
4 tháng 9 2017
M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)
N=\(\sqrt{1+2\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{1+2\left(\sqrt{2}+1\right)}=\) \(\sqrt{1+2\sqrt{2}+2}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))
=\(\sqrt{x}+1+2\sqrt{x}=3\sqrt{x}+1\)
Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )
=\(\left|\sqrt{x}+1\right|+\left|\sqrt{x}-1\right|\)
th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)
th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)
HT
4 tháng 9 2017
a) \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)
\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)
\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)
P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
3 tháng 2 2022
b: \(=\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3+\sqrt{5}+3-\sqrt{5}}{2}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)
c: \(=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
27 tháng 5 2017
\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
27 tháng 5 2017
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
C = \(\sqrt{x-1+2\sqrt{x-1}.1+1}\)+\(\sqrt{1-2\sqrt{x-1}.1+x-1}\)
C = \(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)+ \(\sqrt{\left(1-\sqrt{x-1}\right)^2}\)
C = I \(\sqrt{x-1}+1\)I + I\(1-\sqrt{x-1}\)I
C = \(\sqrt{x-1}+1\) + \(1-\sqrt{x-1}\)
C = 2
\(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(C^2=\left(\sqrt{x+2\sqrt{x-1}}\right)^2+2\sqrt{\left(x+2\sqrt{x-1}\right)\left(x-2\sqrt{x-1}\right)}\) \(+\left(\sqrt{x-2\sqrt{x-1}}\right)^2\)
\(=x+2\sqrt{x-1}+2\sqrt{x^2-\left(2\sqrt{x-1}\right)^2}+x-2\sqrt{x-1}\)
\(=2x+2\sqrt{x^2-4x+1}\)
\(=2\left(x+\sqrt{x^2-4x+1}\right)\)
\(\Rightarrow C=\sqrt{2\left(x+\sqrt{x^2-4x+1}\right)}\)