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a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
\(a,P=\dfrac{-x+2\sqrt{x}-1+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}:\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)
Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2:
a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(2P=2\sqrt{x}+5\)
=>\(P=\sqrt{x}+\dfrac{5}{2}\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)
=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)
=>\(2x+3\sqrt{x}-2=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
Bạn có thể làm hộ mình câu c được không?Nếu được thì mình cảm ơn bạn nhiều!
\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)
\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)
\(\Rightarrow x-y=1\Rightarrow P=1\)
\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)
\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)
\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)
\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)
\(=\sqrt{\left(x^2-x-2\right)\left(x^2-x\right)+5}\)
Đặt \(t=x^2-x\) ta đc:
\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)
\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)
Vậy....
b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=\left|x-2\right|+\left|x+3\right|\)
Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)
Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)
\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7