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24 tháng 8 2017

B1 :

a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)

\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)

b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)

\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)

\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)

\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)

\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)

24 tháng 8 2017

B2 :

a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)

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

b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)

\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)

8 tháng 7 2021

a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)

8 tháng 7 2021

em cảm mơn ạ yeu

4 tháng 7 2021

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

4 tháng 7 2021

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

            \(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)

           \(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)

6 tháng 7 2021

Bài 1:

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

b)Tại x=7 thay vào Q ta được:

\(Q=2.7-\left|7+1\right|=14-8=6\)

Bài 2:

\(\sqrt{x^2-6x}+7x=13\)\(\Leftrightarrow\sqrt{x^2-6x}=13-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-7x\ge0\\x^2-6x=\left(13-7x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\0=48x^2-85x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\\Delta=\left(-85\right)^2-4.48.169=-25223< 0\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm.

6 tháng 7 2021

em cảm mơn nhìu ạ yeu

11 tháng 8 2017

a,\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=|^{ }_{ }2-\sqrt{5}|^{ }_{ }-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)(vì \(2-\sqrt{5}< 0\))

=-2

b,\(\sqrt{16}\cdot\sqrt{25}+\sqrt{256}\cdot\sqrt{64}\)

\(=4\cdot5-16\cdot8=20+128=148\)

c,\(\sqrt{\left(\sqrt{2}-3\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)

\(=|^{ }_{ }\sqrt{2}-3|^{ }_{ }-|^{ }_{ }5-\sqrt{2}|^{ }_{ }\)

\(=3-\sqrt{2}-5+\sqrt{2}\)(vì \(\sqrt{2}-3< 0;5-\sqrt{2}>0\))

\(=-2\)

11 tháng 8 2017

cảm ơn

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

30 tháng 8 2021

các câu ở dưới nữa ah

26 tháng 8 2021

Bài 1: 

\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)

\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)

\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}\right)^2}-\sqrt{2}=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)

Bài 2: 

Ta có: G-1

\(=\dfrac{\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)

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

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ

hay \(G\le1\)

27 tháng 7 2018

B1 

a,\(\frac{5}{3\sqrt{8}}\)\(=\frac{5.\sqrt{8}}{3\sqrt{8}.\sqrt{8}}=\frac{5.\sqrt{8}}{3}\)

b,\(\frac{2a}{1-\sqrt{a}}=\frac{2a\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{2a\left(1+\sqrt{a}\right)}{1-a}\)

c,\(\frac{2+\sqrt{2}}{1+\sqrt{2}}=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)

d,\(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3-1}\right)}{1-\sqrt{3}}=\sqrt{-5}\)

18 tháng 6 2019

a) \(a+b\ge2\sqrt{a}\cdot\sqrt{b}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

b) \(a+b+c\ge\sqrt{a}\cdot\sqrt{b}+\sqrt{a}\cdot\sqrt{c}+\sqrt{b}\cdot\sqrt{c}\)

\(\Leftrightarrow2a+2b+2c-2\sqrt{a}\cdot\sqrt{b}-2\sqrt{a}\cdot\sqrt{c}-2\sqrt{b}\cdot\sqrt{c}\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

18 tháng 6 2019

a)

\(a+b\ge2\sqrt{a}.\sqrt{b}\)

\(\Leftrightarrow\) \(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow\) \(a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow\) \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( vì a, b > 0) luôn đúng

=> Bất đẳng thức đã cho luôn đúng với ∀ a, b dương (đpcm)