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a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
Làm nốt ::v
\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)
\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)
\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)
\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)
\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)
\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)
\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)
Mình làm ba câu mẫu, bạn theo đó mà làm các câu còn lại.
Giải:
1) \(2\sqrt{a^2}\)
\(=2\left|a\right|\)
\(=2a\left(a\ge0\right)\)
Vậy ...
5) \(3\sqrt{9a^6}-6a^3\)
\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)
\(=3.3a^3-6a^3\)
\(=9a^3-6a^3\)
\(=3a^3\)
Vậy ...
10) \(C=\sqrt{4x^2-4x+1}-\sqrt{4x^2+4x+1}\)
\(\Leftrightarrow C=\sqrt{\left(2x-1\right)^2}-\sqrt{\left(2x+1\right)^2}\)
\(\Leftrightarrow C=2x-1^2-\left(2x+1^2\right)\)
\(\Leftrightarrow C=2x-1-2x-1\)
\(\Leftrightarrow C=-2\)
Vậy ...
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
\(1a.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=21-2\sqrt{21}+2\sqrt{21}=21\) \(b.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)
\(2a.\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}.b^2}+\sqrt{\dfrac{a^2}{b^2}.\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+b\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}=\left(2+b\right)\sqrt{\dfrac{a}{b}}\) \(b.\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(x-1\right)^2}}.\sqrt{\dfrac{\left(2\sqrt{m}x-2\sqrt{m}\right)^2}{81}}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{\text{|}2\sqrt{m}x-2\sqrt{m}\text{|}}{9}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{2\sqrt{m}\text{|}x-1\text{|}}{9}=\dfrac{2m}{9}\) \(3a.VP=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1=VT\)
KL : Vậy đẳng thức được chứng minh.
\(b.VP=\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{\text{|}a+b\text{|}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{a+b}=\text{|}a\text{|}=VT\)
KL : Vậy đẳng thức được chứng minh .
P/s : Dài v ~
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
