PT
3
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
21 tháng 7 2017
\(\sqrt{x-1-2\sqrt{x-1}+1}\)+\(\sqrt{x-1+4\sqrt{x-1}+4}\) (\(x\ge1\)
=\(\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}-2\right|\)
dat \(\sqrt{x-1}=t\left(t\ge0\right)\)
ta co \(\left|t-1\right|+\left|t-2\right|\)
t |t-1| |t-2| 1 2 0 0 + - - +
nenta co voi0<= t<1 \(1-t+2-t=3-t=3-2\sqrt{x-1}\)
voi 1\(\le t\le2\) \(t-1+2-t=3\)
voi t>2 \(t-1+t-2=2t-3=2\sqrt{x-1}-3\)
b,\(\sqrt{x-4-4\sqrt{x-4}+4}\) =\(\left|\sqrt{x-4}-2\right|\)
PK
0
TM
2
HC
13 tháng 9 2016
ĐKXĐ: x >= 4
Bình phương hai vế ta có : x + \(\sqrt{x-4}\)+ x - \(\sqrt{x-4}\)
= 2x
26 tháng 6 2019
Theo Bình phương 2 vế ta có :
\(\sqrt{x-4}\)+ X - \(\sqrt{x-4}\)
~ Hok tốt ~
#Gumball
17 tháng 3 2020
B=\(\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{10\sqrt{x}-2x+2\sqrt{x}+3\sqrt{x}-3-x-4\sqrt{x}-\sqrt{x}-4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{-3x+3\sqrt{x}+7\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)
Vậy...
I
17 tháng 3 2020
\(B=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x-3}}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)( \(x\ge0;x\ne1\)
=>\(B=\frac{10\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
=> \(B=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)( zì \(x\ge0,x\ne1\)
D
2
T
10 tháng 9 2018
Đặt\(A=\sqrt{x+\sqrt{x^2-4}}-4.\sqrt{x-\sqrt{x^2-4}}\)
\(A^2=x+\sqrt{x^2-4}+16.\left(x-\sqrt{x^2-4}\right)-2.4.\sqrt{x^2-\left(\sqrt{x^2-4}\right)^2}\)
\(A^2=x+\sqrt{x^2-4}+16x-16.\sqrt{x^2-4}-8.\sqrt{x^2-x^2+4}\)
\(A^2=17x-15.\sqrt{x^2-4}-16\)
mình làm đến đây đc thôi, sorry
11 tháng 9 2018
Dễ thây \(x\ge2\)
\(A=\sqrt{x+\sqrt{x^2-4}}-4\sqrt{x-\sqrt{x^2-4}}\)
\(=\sqrt{\frac{2x+2\sqrt{\left(x+2\right)\left(x-2\right)}}{2}}-4\sqrt{\frac{2x-2\sqrt{\left(x+2\right)\left(x-2\right)}}{2}}\)
\(=\sqrt{\frac{\left(x+2\right)+2\sqrt{\left(x+2\right)\left(x-2\right)}+\left(x-2\right)}{2}}-4\sqrt{\frac{\left(x+2\right)-2\sqrt{\left(x+2\right)\left(x-2\right)}+\left(x-2\right)}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{\left(x+2\right)}+\sqrt{\left(x-2\right)}\right)^2}{2}}-4\sqrt{\frac{\left(\sqrt{\left(x+2\right)}-\sqrt{\left(x-2\right)}\right)^2}{2}}\)
\(=\frac{1}{\sqrt{2}}\left[\left(\sqrt{x+2}+\sqrt{x-2}\right)-4\left(\sqrt{x+2}-\sqrt{x-2}\right)\right]\)
\(=\frac{1}{\sqrt{2}}\left(-3\sqrt{x+2}+5\sqrt{x-2}\right)\)
cách khác nhé:
ĐK: \(x\ge4\)
\(B=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{\left(x-4\right)+4\sqrt{x-4}+4}+\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
Nếu \(4\le x< 8\)thì: \(B=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Nếu \(x\ge8\)thì: \(B=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
\(B=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(\Leftrightarrow B^2=\left(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\right)^2\)
\(=x+4\sqrt{x-4}+x-4\sqrt{x-4}+2\sqrt{\left(x+4\sqrt{x-4}\right)\left(x-4\sqrt{x-4}\right)}\)
\(=2x+2\sqrt{x^2-\left(4\sqrt{x-4}\right)^2}\)
\(=2x+2\sqrt{x^2-16\left(x-4\right)}=2x+2\sqrt{x^2-16x+64}\)
\(=2x+2\sqrt{\left(x-8\right)^2}=2x+2\left|x-8\right|\)
Nếu \(x-8\ge0\Rightarrow x\ge8\) thì 2x + 2(x-8) = 2x + 2x - 16 = 4x -16 = 4(x-4)
Nếu x - 8 < 0 => x < 8 thì 2x + 2(8 - x) = 2x + 16 - 2x = 0x + 16