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a: ĐKXĐ: x>=0; x<>1
b: Khi x=9 thì \(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{x-1}\)
\(=\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)^2}=\dfrac{6}{\sqrt{x}+1}\)
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 \(x=3\)
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)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
a: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
bài 2 ) a) đk : \(a>0;b>0\)
b) P = \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
P = \(\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
P = \(\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) = \(a-b\)
c) ta có P = \(a-b\) thay \(a=2\sqrt{3};b=\sqrt{3}\) vào ta có
P = \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\) vậy khi \(a=2\sqrt{3};b=\sqrt{3}\) thì P = \(\sqrt{3}\)
bài 1) a) P = \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
P = \(\dfrac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-a^2\sqrt{a}+a^2-a+\sqrt{a}}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(\dfrac{2a^2-2a}{a^2-a}+\dfrac{2a+1}{\sqrt{a}}\) = \(\dfrac{2\left(a^2-a\right)}{a^2-a}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(2+\dfrac{2a+2}{\sqrt{a}}\) = \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b) ta có P = 7 \(\Leftrightarrow\) \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\) \(\Leftrightarrow\) \(2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow\) \(2a-5\sqrt{a}+2=0\) (1)
đặc \(\sqrt{a}=u\) \(\left(u\ge0\right)\) (1) \(\Leftrightarrow\) \(2u^2-5u+2\)
\(\Delta=\left(-5\right)^2-4.2.2\) = \(25-16=9>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(u_1=\dfrac{5+3}{4}=\dfrac{8}{4}=2\left(tmđk\right)\)
\(u_2=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\left(tmđk\right)\)
ta có : \(u=\sqrt{a}=2\Leftrightarrow x=4\)
\(u=\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(a=4;a=\dfrac{1}{4}\) thì P = 7
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
Bài 1:
a: \(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{x-9}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b: Để A=3/4 thì căn x+2=3
=>x=1
c: Khi x=4 thì \(A=\dfrac{2+2}{2+3}=\dfrac{4}{5}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)
c: Ta có: P=AB
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}+1}\)