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a, \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
b, Với a;b > 0
\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
c, Với x >= 0
\(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d, Với x >= 0 ; x khác 14
\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\frac{1}{\sqrt{3}}\)
b) \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)
c) \(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{\left(4\sqrt{x}+7\right)}=\sqrt{x}-1\)
d) \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
ĐKXĐ : \(x\ne1;x\ne0;x>0\)
a) \(A=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\cdot\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{x-1}{\sqrt{x}}\right)\)
\(=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4x\sqrt{x}-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{4x\sqrt{x}}{\sqrt{x}}=4x\)
b) \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}\)
Suy ra : \(A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}\)
c) \(\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x\)
\(\Leftrightarrow2\sqrt{x}>4x\)
\(\Leftrightarrow2\sqrt{x}-4x>0\)
\(\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0\)
\(\Rightarrow1-2\sqrt{x}>0\) (do x > 0 nên \(2\sqrt{x}>0\))
\(\Leftrightarrow1>2\sqrt{x}\)
\(\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}\)
Theo điều kiện suy ra giá trị của x để \(\sqrt{A}>A\) là \(0< x< \dfrac{1}{4}\)
a) \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{x-1}{\sqrt{x}}\)
\(=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=4\sqrt{x}\cdot\left[1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]\cdot\dfrac{1}{\sqrt{x}}\)
\(=4\left(1+x-1\right)\)
\(=4x\)
b) Thay \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) vào biểu thức A.
Ta có:
\(4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\\ =-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12\)
Vậy giá trị biểu thức A tại \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) là \(-4\sqrt{6}+12\)
c) Để \(\sqrt{A}>A\)
\(\Rightarrow\sqrt{4x}>4x\)
\(\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)\)
a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{x-\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) thì
\(P=\dfrac{\sqrt{3}-1+1}{\sqrt{3}-1}=\dfrac{\sqrt{3}}{\sqrt{3}-1}\)
1) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-9\ne0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}\)2) Để A=\(\dfrac{5}{6}\) thì \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}=\dfrac{5}{6}\Leftrightarrow\left(\sqrt{x}+1\right)6=\left(\sqrt{x}+3\right)5\Leftrightarrow6\sqrt{x}+6=5\sqrt{x}+15\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)
1. Ta có:
\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)}{3\left(x-9\right)}+\dfrac{1}{3}\)
\(=\dfrac{2x-6\sqrt{x}}{3\left(x-9\right)}+\dfrac{x-9}{3\left(x-9\right)}\)
\(=\dfrac{3x-6\sqrt{x}-9}{3x-27}\)
\(=\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{8\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+8}{\sqrt{x}-3}\)
Do \(A>0\) \(\forall x\ge0\Rightarrow\)để P xác định thì \(B\ge0\Rightarrow x>9\)
\(\Rightarrow P=\sqrt{\dfrac{\sqrt{x}+8}{\sqrt{x}-3}.\dfrac{x+7}{\sqrt{x}+8}}=\sqrt{\dfrac{x+7}{\sqrt{x}-3}}=\sqrt{\sqrt{x}+3+\dfrac{16}{\sqrt{x}-3}}\)
\(\Rightarrow P=\sqrt{\sqrt{x}-3+\dfrac{16}{\sqrt{x}-3}+6}\ge\sqrt{2\sqrt{\dfrac{16\left(\sqrt{x}-3\right)}{\sqrt{x}-3}}+6}=\sqrt{14}\)
\(\Rightarrow P_{min}=\sqrt{14}\) khi \(x=49\)
Ta có :
\(A.B=\dfrac{24}{\sqrt{x}+6}.\dfrac{\sqrt{x}+6}{\sqrt{x}-6}\)
\(=\dfrac{24}{\sqrt{x}-6}\)
Để \(AB\le12\Leftrightarrow\dfrac{24}{\sqrt{x}-6}\le12\)
\(\Leftrightarrow\dfrac{24-12\left(\sqrt{x}-6\right)}{\sqrt{x}-6}\le0\)
\(\Leftrightarrow24-12\sqrt{x}+72\le0\)
\(\Leftrightarrow-12\sqrt{x}\le-96\)
\(\Leftrightarrow\sqrt{x}\ge8\)
\(\Leftrightarrow x\ge64\)
Vậy \(x\ge64\) thì \(AB\le12\)