\(\sqrt{8}\)- 4\(\sqrt{2}+\sqrt{40}\)...">
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13 tháng 8 2018

a, \(M=\left(\sqrt{8}-4\sqrt{2}+\sqrt{40}\right)\sqrt{2} \)
\(=\sqrt{16}-4\sqrt{4}+\sqrt{80}\\ =4-8+\sqrt{16.5}\\ =4\sqrt{5}-4\)
\(N=\dfrac{1}{\sqrt{5}+2}=\dfrac{\sqrt{5}-2}{5-4}=\sqrt{5-2}\)

b, \(M+N=4\sqrt{5}-4+\sqrt{5}-2=5\sqrt{5}-6\)

30 tháng 6 2018

có phải/....

1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)

2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

30 tháng 6 2018

1.B=\(\dfrac{\sqrt{x-1}}{\sqrt{x+2}}\)

AH
Akai Haruma
Giáo viên
14 tháng 5 2018

Lời giải:

a)

Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)

Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)

\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)

\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)

\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)

b)

ĐK: \(x\geq 0; x\neq 4\)

\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)

\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)

\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

 

 

AH
Akai Haruma
Giáo viên
14 tháng 5 2018

c) Thêm ĐK: \(x\geq 1\)

Từ biểu thức P vừa tìm được:

\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

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

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

\((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)

\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :

\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)

Vậy..........

1 tháng 10 2021

\(\sqrt{x+2\sqrt{x+1}}\)

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

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

\(\left|\sqrt{x-1}+1\right|\)

1 tháng 10 2021

a )  Với \(x\ge1\) ta có : 

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

b ) Với \(x\ge\frac{1}{2}\) ta có : \(N=\sqrt{2x-1+4\sqrt{2x-1}+4}=\sqrt{\left(\sqrt{2x-1}+2\right)^2}=\sqrt{2x-1}+2\)

a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)

b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)

c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)

d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)

e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)

f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)

8 tháng 8 2018

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)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)

8 tháng 8 2018

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 : \(0< a< 1\)

16 tháng 6 2017

đk : \(x\ne4\) ; \(x\ge0\)

1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)

Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)

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

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

Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

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

Q = \(\dfrac{3}{2+\sqrt{x}}\)

b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)

\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)

c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4

ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên

\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3

\(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3

\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)

vậy x = 1 nguyên thì Q nguyên

16 tháng 6 2017

2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)

= \(4\sqrt{a}-5\sqrt{10a}\)

b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)

c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

= \(33-2\sqrt{22}-11+3\sqrt{22}\)

= \(22+\sqrt{22}\)

Bài 1: 

a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)

b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)

c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)