Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)
\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)(1)
b) Ta có: \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào biểu thức (1), ta được:
\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)
Tìm được A = 24 5 và B = - 6 x - 4 với x > 0 và x ≠ 4 ta tìm được 0 < x < 1
Ta có M = - 1 + 2 x ∈ Z => x ∈ Ư(2) từ đó tìm được x=1
a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)
c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì
\(M=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-2}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}\)
c: M>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\)
mà \(\sqrt{x}>0\)
nên \(\sqrt{x}-2>0\)
=>\(\sqrt{x}>2\)
=>x>4
a: Ta có: \(x^2=3-2\sqrt{2}\)
nên \(x=\sqrt{2}-1\)
Thay \(x=\sqrt{2}-1\) vào A, ta được:
\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)
a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
1: \(1+\sqrt{\dfrac{\left(x-1\right)^2}{x-1}}=1+\sqrt{x-1}\)
2: \(A=\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)
=\(\left|x-2\right|+\dfrac{x-2}{\left|x-2\right|}\)
TH1: x>2
A=x-2+(x-2)/(x-2)=x-2+1=x-1
TH2: x<2
A=2-x+(x-2)/(2-x)=2-x-1=1-x
3: \(C=\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)
\(=\sqrt{m}-\sqrt{\left(\sqrt{m}-1\right)^2}\)
\(=\sqrt{m}-\left|\sqrt{m}-1\right|\)
TH1: m>=1
\(C=\sqrt{m}-\sqrt{m}+1=1\)
TH2: 0<=m<1
\(C=\sqrt{m}+\sqrt{m}-1=2\sqrt{m}-1\)
a, V = 1 x + 2 + 1 x − 2 x + 2 x = x − 2 + x + 2 x + 2 x − 2 x + 2 x = 2 x − 2
b, V = 1 3 ⇔ 2 x − 2 = 1 3 ⇔ x − 2 = 6 ⇔ x = 64 ( t / m )
b: A=x+|x-2|
TH1: x>=2
A=x+x-2=2x-2
TH2: x<2
A=x+2-x=2
c: B=|x-3|-x
TH1: x>=3
B=x-3-x=-3
TH2: x<3
B=3-x-x=3-2x