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\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Do \(a\ge0\)
\(\Rightarrow a\in\left\{0;2;3\right\}\)
Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)
\(=\dfrac{2}{a-1}\)
Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{2;0;3\right\}\)
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
Đăt\(\sqrt{a}\)=x, \(\sqrt{b}\)=y (x,y>0)
=>xy+1=4y => 4y≥ \(2\sqrt{xy}\)=>\(2\sqrt{y}\)≥\(\sqrt{x}\)=> 4y≥x=> 4≥ \(\dfrac{x}{y}\)=> \(\dfrac{1}{4}\)≤\(\dfrac{y}{x}\)=>\(\dfrac{-1}{4}\)≥\(\dfrac{-y}{x}\)
Xét:A=(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{y-x}\)+1):(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{x-y}\)-1)
= \(\dfrac{-2y^2\left(x+1\right)}{\left(x-y\right)\left(x+y\right)}\).\(\dfrac{\left(x-y\right)\left(x+y\right)}{2xy\left(x+1\right)}\)
=> A= \(\dfrac{-y}{x}\)≤\(\dfrac{-1}{4}\)
Dấu "=" xảy ra <=> xy=1 và x=4y <=> x=2, y=\(\dfrac{1}{2}\) <=> a =4, b=\(\dfrac{1}{4}\)
Vậy Max A =\(\dfrac{-1}{4}\) <=> a=4, b=\(\dfrac{1}{4}\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
a: \(P=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{a-b}+b\right]\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)}+b\right]\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)}{a-\sqrt{ab}+b}\cdot\left(\dfrac{\sqrt{a}}{a+\sqrt{ab}+b}+b\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{\sqrt{a}+ab+b\sqrt{ab}+b^2}{a+\sqrt{ab}+b}\)
b: Thay a=16 và b=4 vào P, ta được:
\(P=\dfrac{4+2}{16-4\cdot2+4}\cdot\dfrac{4+16\cdot4+4\cdot4\cdot2+16}{16+4\cdot2+4}\)
\(=\dfrac{6}{12}\cdot\dfrac{116}{28}=\dfrac{29}{14}\)
Câu (A) đề có sao không nhỉ?
\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow\dfrac{1}{a-1}\)
\(E=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)
a: \(P=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{a\sqrt{a}-b\sqrt{b}}:\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2\left(a+\sqrt{ab}+b\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=2/a-1
b: Để P nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{2;0;3\right\}\)