Cho D= (\(\dfrac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\sqrt{\dfrac{\sqrt{a}+\sqrt{b}}{1+\sqrt{ab}}}\)):\(\left(1+\dfrac{a+b+2ab}{1-ab}\right)\)
a) Rút gọn D
b) TÌm GTLN của D
K
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13 tháng 6 2018
Mạn phép ko chép lại đề , mk làm luôn
a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)
\(D=\dfrac{2\left(\sqrt{a}+\sqrt{b}\right)}{\left(b+1\right)\left(a+1\right)}\)
13 tháng 6 2018
D=A/B
a)
B=1+(a+b+2ab)/(1-ab)=(a+b+ab)/(1-ab)
dk: a,b≥0; a.b≠1
1/B=(1-ab)/(a+b+ab)
A=√a+√b)[(1+√ab)+(1-√ab)]/(1-ab)=2(√a+√b)/(1-ab)
D=2(√a+√b)/[(a+1)(b+1)]
b)
a=2/(√3+2)=2(2-√3)/[(2+√3)(2-√3)]=2(2-√3)=(√3-1)^2
15 tháng 11 2018
1, ĐKXĐ: x\(\ge0\);x\(\ne1\)
Rút gọn P với \(x\ge0;x\ne1\)ta có
P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\dfrac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
15 tháng 11 2018
2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:
P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)
=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)
=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)
\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3
HP
3 tháng 9 2021
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
3 tháng 9 2021
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên
B
11 tháng 7 2023
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=\(\dfrac{1}{4}\)
16 tháng 9 2018
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\) \(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\) \(\Leftrightarrow P=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)\(\Leftrightarrow P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) \(x>0\Rightarrow-\sqrt{x}< 0\) và \(x< 1\Rightarrow\sqrt{x}-1< 0\)
\(\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\) (đpcm)
c) ta có : \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow P_{max}=\dfrac{1}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
vậy GTLN của \(P\) là \(\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)