rút gọn \(B=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
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0
6 tháng 2 2022
Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) − \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)
=2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\)
=(2− 3+ 8)\(\sqrt{2}\)
=7\(\sqrt{2}\)
Câu 2: Mik ko chắc làm đúng hay ko 
nên ko làm
9 tháng 3 2022
Với x >= 0 ; x khác 16
\(B=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
11 tháng 5 2022
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)
\(=\dfrac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}}\)
=6
19 tháng 6 2023
a: \(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{1}{\sqrt{x}-2}\)
b: Khi x=9 thì B=1/(3-2)=1
20 tháng 8 2021
a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)
\(=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)+\sqrt{x}\)
\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}+\sqrt{x}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}+\sqrt{x}\)
\(=-\sqrt{x}-1+\sqrt{x}\)
=-1
AD
23 tháng 7 2023
\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)
\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\) (vì \(x>4\) )
@seven
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,\(\sqrt{x-4-4\sqrt{x-4}+4}\) =\(\left|\sqrt{x-4}-2\right|\)
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