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27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
TN
8 tháng 7 2017
a)\(\sqrt{x^2-2x+1}-\sqrt{x^2-4x+4}=x-3\)
\(\Leftrightarrow\left(\sqrt{x^2-2x+1}-3\right)-\left(\sqrt{x^2-4x+4}-2\right)=x-3-1\)
\(\Leftrightarrow\frac{x^2-2x+1-9}{\sqrt{x^2-2x+1}+3}-\frac{x^2-4x+4-4}{\sqrt{x^2-4x+4}+2}=x-4\)
\(\Leftrightarrow\frac{x^2-2x-8}{\sqrt{x^2-2x+1}+3}-\frac{x^2-4x}{\sqrt{x^2-4x+4}+2}-\left(x-4\right)=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-4\right)}{\sqrt{x^2-2x+1}+3}-\frac{x\left(x-4\right)}{\sqrt{x^2-4x+4}+2}-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+2}{\sqrt{x^2-2x+1}+3}-\frac{x}{\sqrt{x^2-4x+4}+2}-1\right)=0\)
Dễ thấy: \(\frac{x+2}{\sqrt{x^2-2x+1}+3}-\frac{x}{\sqrt{x^2-4x+4}+2}-1< 0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=1\)
\(\Leftrightarrow\left(\sqrt{x^2-6x+9}-\frac{7}{2}\right)-\left(\sqrt{x^2+6x+9}-\frac{5}{2}\right)=0\)
\(\Leftrightarrow\frac{x^2-6x+9-\frac{49}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{x^2+6x+9-\frac{25}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\frac{\frac{4x^2-24x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{4x^2+24x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\frac{\frac{\left(2x-13\right)\left(2x+1\right)}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{\left(2x+1\right)\left(2x+11\right)}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{\frac{2x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{2x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}\right)=0\)
Dễ thấy: \(\frac{\frac{2x-13}{4}}{\sqrt{x^2-6x+9}+\frac{7}{2}}-\frac{\frac{2x+11}{4}}{\sqrt{x^2+6x+9}+\frac{5}{2}}< 0\)
\(\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
c)Áp dụng BĐT CAuchy-Schwarz ta có:
\(P^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\)
\(\le\left(1+1\right)\left(x-2+4-x\right)\)
\(=2\cdot\left(x-2+4-x\right)=2\cdot2=4\)
\(\Rightarrow P^2\le4\Rightarrow P\le2\)
16 tháng 10 2016
\(\frac{2}{P}+\sqrt{x}=\frac{-2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}+\sqrt{x}\)
= - \(2\sqrt{x}-2-\frac{2}{\sqrt{x}}+\sqrt{x}\)
= \(\frac{-2}{\sqrt{x}}-2-\sqrt{x}\le-2-2\sqrt{2}\)
Đạt được khi x = 2
3 tháng 9 2022
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
TB
14 tháng 8 2018
a, \(B=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)(ĐK: \(x\ne1\))
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
\(=\frac{x-2\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b, ĐK: \(x\ne1\)
\(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\sqrt{5}+2-\sqrt{5}+2=4\)
Thay \(x=4\left(TM\right)\)vào B ta có:
\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)
Vậy với \(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)nên \(B=\frac{1}{3}\)
c. ĐK: \(x\ne1\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\)\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le2\Leftrightarrow\frac{-2}{\sqrt{x}+1}\ge-2\)\(\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge-1\)
Dấu = xảy ra \(\Leftrightarrow\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy \(MinB=-1\Leftrightarrow x=0\)
d, ĐK: \(x\ne1\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(B\inℤ\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\inℤ\Leftrightarrow\frac{2}{\sqrt{x}+1}\inℤ\)\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\Leftrightarrow\sqrt{x}+1\in\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy với \(x=0\)thì \(B\inℤ\)