A= x-2 2+ sqrt x (x>=0); ==( 8x sqrt x -1 2x- sqrt x - 8x sqrt x +1 2x+ sqrt x )= 2x+1 2x-1 vdi x>0,x ne 1 2 ;x ne- 1 2 MS05. Cho A =- a. Rút gọn B. b. Tim x d hat e A B =1
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PN
7 tháng 10 2023
\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
PN
7 tháng 10 2023
\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)
29 tháng 7 2023
a: \(A=\dfrac{x\sqrt{2}}{x\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{x-2}\)
\(=\dfrac{1}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+\sqrt{2}}\)
b: \(M=\left(\dfrac{\sqrt{a}+a}{\sqrt{a}-2}\right)\cdot\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\left(\sqrt{a}-2\right)=\sqrt{a}\left(\sqrt{a}-2\right)\)
29 tháng 7 2023
\(A=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)
\(=\dfrac{\sqrt{x}.\sqrt{2x}}{\sqrt{2x}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}=1\)
\(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{\sqrt{a}+a}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 8 2023
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
14 tháng 10 2021
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
TP
Thảo Phương
CTVVIP
21 tháng 6 2019
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
TP
Thảo Phương
CTVVIP
21 tháng 6 2019
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
29 tháng 8 2021
6: Để P>1 thì P-1>0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}>0\)
\(\Leftrightarrow\sqrt{a}-2< 0\)
hay a<4
Kết hợp ĐKXĐ, ta được: \(0\le a< 4\)
30 tháng 8 2021
5: Để P>0 thì \(x-4\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}-4>0\)
hay x>16
VT
1
1 tháng 9 2020
Ta có: \(\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}\)
\(=\frac{\sqrt{a+x^2-2x\sqrt{a}}-\sqrt{a+x^2+2\cdot x\cdot\sqrt{a}}}{\sqrt{x}}\)
\(=\frac{\sqrt{\left(x-\sqrt{a}\right)^2}-\sqrt{\left(x+\sqrt{a}\right)^2}}{\sqrt{x}}\)
\(=\frac{\left|x-\sqrt{a}\right|-\left|x+\sqrt{a}\right|}{\sqrt{x}}\)
\(=\frac{\left|x-\sqrt{a}\right|-x-\sqrt{a}}{\sqrt{x}}\)
\(=\left[{}\begin{matrix}\frac{x-\sqrt{a}-x-\sqrt{a}}{\sqrt{x}}\left(x\ge\sqrt{a}\right)\\\frac{\sqrt{a}-x-x-\sqrt{a}}{\sqrt{x}}\left(x< \sqrt{a}\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}-\frac{2\sqrt{a}}{\sqrt{x}}\\-2\sqrt{x}\end{matrix}\right.\)
16 tháng 5 2023
2:
a: =>2x^2-4x-2=x^2-x-2
=>x^2-3x=0
=>x=0(loại) hoặc x=3
b: =>(x+1)(x+4)<0
=>-4<x<-1
d: =>x^2-2x-7=-x^2+6x-4
=>2x^2-8x-3=0
=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)
