Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(A=2\left(3x+1\right)=2\cdot\left(-3\sqrt{2}+1\right)=-6\sqrt{2}+2\)
a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)
b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))
\(A=\frac{3\left(x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{3x+3\sqrt{x}-3-x+2\sqrt{x}-1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
b) \(x=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)
\(\sqrt{x}=1+\sqrt{2}\)
ý b tự thay vào nha
Bài 1:
a) \(B=\sqrt{1-4x+4x^2}\)
\(=\sqrt{\left(1-2x\right)^2}\)
\(=\left|1-2x\right|\)
Nếu \(x\le\frac{1}{2}\)thì: \(B=1-2x\)
Nếu \(x>\frac{1}{2}\)thì: \(B=2x-1\)
b) Tại \(x=-7\)thì: \(B=1-2.\left(-7\right)=15\)
Lời giải :
a) \(\sqrt{x^2\left(x-1\right)^2}=\left|x\right|\cdot\left|x-1\right|=-x\left(1-x\right)=x^2-x\)
b) \(\sqrt{13x}\cdot\sqrt{\frac{52}{x}}=\sqrt{\frac{13x\cdot52}{x}}=\sqrt{676}=26\)
c) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)