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Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
Giải:
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)
\(=\sqrt{2}\)
Vậy ...
a) Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=\dfrac{-\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{-\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)
\(=-2\sqrt{2}\)
b) Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
c) Ta có: \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
\(=\left(\dfrac{-\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}-2\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right)\)
\(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
\(=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=-1\)
d) Ta có: \(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(=\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
ĐKXĐ: x>=0; x<>1
a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)
c: B=2/3
=>căn x/căn x+1=2/3
=>căn x=2
=>x=4
d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)
=>B<1
e: B>1
=>-1/căn x+1>0
=>căn x+1<0(vô lý)
=>KO có x thỏa mãn
f: B nguyên khi căn x chia hết cho căn x+1
=>căn x+1-1 chia hết cho căn x+1
=>căn x+1=1 hoặc căn x+1=-1(loại)
=>căn x=0
=>x=0
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
`c)(15/(sqrt6+1)+4/(sqrt6-2)-12/(3-sqrt6))*(sqrt6+11)`
`=((15(sqrt6-1))/(6-1)+(4(sqrt6+2))/(6-4)-(12(3+sqrt6))/(9-6))*(sqrt6+11)`
`=(3(sqrt6-1)+2(sqrt6+2)-4(3+sqrt6))*(sqrt6+11)`
`=(3sqrt6-3+2sqrt6+4-12-4sqrt6)*(sqrt6+11)`
`=(sqrt6-11)(sqrt6+11)`
`=6-121=-115`
c) Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121=-115
a) Ta có: \(P=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
\(=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)
\(=\left(\dfrac{x+2\sqrt{xy}+y}{xy}\right):\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
a) Đk:\(x>0;y>0\)
\(P=\left[\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}.\sqrt{y}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{x\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{xy}+y\sqrt{xy}}\)
\(=\left[\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\right]:\dfrac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(x+y\right)}\)
\(=\dfrac{2\sqrt{xy}+x+y}{xy}:\dfrac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy}.\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
b) \(xy=16\Leftrightarrow x=\dfrac{16}{y}\)
\(P=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=\dfrac{1}{\sqrt{\dfrac{16}{y}}}+\dfrac{1}{\sqrt{y}}=\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\)
Áp dụng AM-GM có:
\(\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\ge2\sqrt{\dfrac{\sqrt{y}}{4}.\dfrac{1}{\sqrt{y}}}=1\)
\(\Rightarrow P\ge1\)
Dấu "=" xảy ra khi \(y=4\Rightarrow x=4\)
Vậy x=y=4 thì P đạt GTNN là 1
@Azue help me
help me