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HT
25 tháng 6 2018
a, \(\left(\sqrt{x-1}-2\right)^2+\)\(\left(\sqrt{x-1}-3\right)^2\)
xog xét 2 TH
b, bình phương
2
GTLN : 2 dấu = xra \(2\le x\le4\)
18 tháng 11 2019
a) \(x\ge0\)đặt \(\sqrt{x}=a\ge0\)
\(A=\frac{2a}{a^2-a+1}\Leftrightarrow A.a^2+A-2a=0\Leftrightarrow A.a^2-\left(A+2\right)a+A=0\)
\(\Delta=\left(A+2\right)^2-4A^2=-3A^2+4A+4\ge0\Rightarrow A\le2\)
\(\Rightarrow A_{max}=2\) khi \(x=1\)
b)
\(x\ge0\)
\(B=-\left(x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)-\frac{7}{4}=-\left(\sqrt{x-\frac{1}{2}}\right)^2-\frac{7}{4}\le\frac{-7}{4}\)
\(\Rightarrow B_{max}=\frac{-7}{4}\) khi \(\sqrt{x=}\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
c) \(x\ge0\)
\(C=-2+\sqrt{x}-1=-2\left(x-2.\sqrt{x}.\frac{1}{4}+\frac{1}{16}\right)-\frac{7}{8}\)
\(C=-2\left(\sqrt{x}-\frac{1}{4}\right)^2\frac{7}{8}\le\frac{-7}{8}\)
\(C_{max}=\frac{-7}{8}\)khi đó \(x=\frac{1}{16}\)
27 tháng 7 2017
1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ : \(2\le x\le4\)
\(\Rightarrow A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
Áp dụng bđt AM - GM ta có :
\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)
\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)
Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2
=> A = \(\sqrt{2}\)
Vậy \(\sqrt{2}\le A\le2\)
13 tháng 9 2016
a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
19 tháng 12 2018
ĐK: x\(\ge0,x\ne1\)
a) \(Q=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Ta có \(Q=0,5\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=0,5\Leftrightarrow2-5\sqrt{x}=0,5\sqrt{x}+1,5\Leftrightarrow0,5=5,5\sqrt{x}\Leftrightarrow\sqrt{x}=\dfrac{1}{11}\Leftrightarrow x=\dfrac{1}{121}\left(tm\right)\)
Vậy \(x=\dfrac{1}{121}\) thì \(Q=0,5\)
c) Ta có \(Q=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}=\dfrac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)
Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}+\left(-5\right)\le\dfrac{2}{3}\Leftrightarrow\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{2}{3}\Leftrightarrow Q\le\dfrac{2}{3}\)
Dấu bằng xảy ra khi x=0
Vậy GTLN của Q=\(\dfrac{2}{3}\)
6 tháng 8 2016
a) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{\left(2\sqrt{x}-2\right)-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right]\left(ĐK:x\ge0;x\ne9\right)\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{-3}{\sqrt{x}+3}\)
\(B=\sqrt{x-2}+\sqrt{11-x}\)
\(=\sqrt{x-2}.1+\sqrt{11-x}.1\)
\(\le\sqrt{\left|\left(x-2+11-x\right)\left(1+1\right)\right|}=3\sqrt{2}\)
Dấu \(=\)khi \(x-2=11-x\Leftrightarrow x=\frac{13}{2}\).
ĐKXĐ:\(2\le x\le11\)
Có:\(B=\sqrt{x-2}+\sqrt{11-x}\)
Áp dụng bđt B.C.S ta có:
\(|\sqrt{x-2}+\sqrt{11-x}|\le\sqrt{\left(1+1\right)\left(x-2+11-x\right)}\)
\(\Leftrightarrow|\sqrt{x-2}+\sqrt{11-x}|\le3\sqrt{2}\)
Mà\(\sqrt{x-2}+\sqrt{11-x}\le\left|\sqrt{x-2}+\sqrt{11-x}\right|\)
\(\sqrt{x-2}+\sqrt{11-x}\le3\sqrt{2}\)
Hay\(B\le3\sqrt{2}\)
Dấu "=" xảy ra <=>\(x-2=11-x\)
\(\Leftrightarrow x=\frac{13}{2}\)(Thỏa mãn ĐKXĐ)
Vậy Max\(B=3\sqrt{2}\)\(\Leftrightarrow x=\frac{13}{2}\)