\(\sqrt{2a}\) \(\times\)
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10 tháng 7 2018

\(a.\sqrt{2a}.\sqrt{18a}=\sqrt{2a}.3\sqrt{2a}=3.2a=6a\)

\(b.\sqrt{3a.27ab^2}=\sqrt{9a^2b^2.9}=9\text{ |}ab\text{ |}\)

\(c.2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^2}{-2y}=-x^2y\)

\(d.\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)

\(e.\sqrt{\dfrac{9a^2}{16}}=\dfrac{3\text{ |}a\text{ |}}{4}\)

\(f.\sqrt{10.16a^2}=-4a\sqrt{10}\)

\(g.\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)

\(h.\sqrt{\dfrac{2a^2b^4}{98}}\sqrt{\dfrac{a^2b^4}{49}}=\dfrac{b^2\text{ |}a\text{ |}}{7}\)

17 tháng 8 2018

\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)

\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)

\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)

\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)

\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)

\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)

2 tháng 9 2018

Đề câu c co bị sai ko vậy bạn? (y - 2\(\sqrt{x}\) +1)

a: \(=\sqrt{3}+1-\sqrt{3}=1\)

b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)

c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)

19 tháng 8 2018

B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)

=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)

=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)

=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)

b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)

\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)

e: \(=2-x-x=2-2x\)

i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)

\(=\left|2x-1\right|-2x\)

=1-2x-2x=1-4x

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: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)

\(=\dfrac{x^2}{y}\cdot y-x^2=0\)

 

13 tháng 7 2018

Biết đâu làm đó , sai thôi đừngg chửi nhé

1, Rút gọn

a) A = \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\) = \(\dfrac{\left(\sqrt{x}\right)^2+\sqrt{xy}}{\left(\sqrt{y}\right)^2+\sqrt{xy}}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}\) = \(\dfrac{\sqrt{x}}{\sqrt{y}}\)

b) B = \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) . \(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\)

= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}.\dfrac{bc^3}{\left(a-b\right)}}\) = \(\sqrt{\left(a-b\right)^2.b^4.c^2}\)

= \(\left|a-b\right|\) . \(\left|b^2\right|\) . \(\left|c\right|\)

= -(a -b) .b2. c

13 tháng 7 2018

bài 2:

a/ \(\sqrt{x^2-4}-\sqrt{x-2}=0\) đk: x≥2

<=> \(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

<=>\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

vậy pt có 1 nghiệm x = 2

b/ \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)

Ta có: \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=\sqrt{3\left(x^2+4x+4\right)+4}+\sqrt{\left(y^2-4y+4\right)+9}=\sqrt{3\left(x+2\right)^2+4}+\sqrt{\left(y-2\right)^2+9}\ge\sqrt{4}+\sqrt{9}=2+3=5\)

=> Dấu ''='' xảy ra khi x = -2; y = 2

Vậy pt có nghiệm x=-2; y = 2

26 tháng 6 2018

Làm nốt ::v

\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)

\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)

\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)

\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)

\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)

\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)

\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)

25 tháng 6 2018

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 ...

a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)

b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)

\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)

c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)

5 tháng 8 2018

\(A=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

5 tháng 7 2018

\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)

\(\Leftrightarrow3< 1\) ( Vô lý )

\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)

\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)

\(\Leftrightarrow2b-2\sqrt{ab}< 0\)

\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)

Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)

\(\RightarrowĐpcm.\)

\(2a.\) Áp dụng BĐT Cauchy , ta có :

\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)

\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)

\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)

Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :

\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)

5 tháng 7 2018

\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)

\(\Leftrightarrow x-4=a^2\)

\(\Leftrightarrow x=a^2+4\left(TM\right)\)

\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)

\(\Leftrightarrow x+4=x^2+4x+4\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)

KL....