Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)
\(A=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=4\sqrt{x}-\left(\sqrt{x}+3\right)\)
\(=3\sqrt{x}-3\)
\(B=\frac{\sqrt{\left(3x+2\right)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)
\(TH1:3x+2>0\Rightarrow B=1\)
\(TH2:3x+2< 0\Rightarrow B=-1\)
A <=> 4√x - [ ( (√x )^2 + 2√x3+ 3^2)*( √x -3)]/ (x-9)
<=> 4√x - [(√x+3)^2×(√x-3)]/( x-9)
<=> 4√x - [(√x+3)*(x-9)]/(x-9)
<=> 4√x - √x -3
<=> 3√x -3
b, <=> √[(3*x) ^2+2*3x*2+2^2]/(3x+2)
<=> √[( 3x+2)^2] /(3x+2)
<=> (3x+2)/(3x+2) = 1
\(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\)
\(=\left(\frac{3x-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{3x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=3\sqrt{x}\)
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
Em mới học lớp 8 nhưng làm thử sai thì thôi nhé !!!
\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}\)\(-\frac{3x+3}{x-9}\)
\(P=\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\sqrt{x}^2-3^2}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\sqrt{x}^2-3^2}\)
\(p=\frac{-3\sqrt{x}-3}{\sqrt{x}^2-3^2}=\frac{-3.\left(\sqrt{x}+1\right)}{x-9}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé