K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a: \(VT=\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)

\(=\dfrac{-\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}=\dfrac{-3\sqrt{6}+4\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)

b: \(VT=\dfrac{\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)}{\sqrt{6x}}\)

\(=1+\dfrac{1}{3}+1=2\dfrac{1}{3}\)

23 tháng 5 2021

a) -17√3/3                                                  b) 11√6 

c) 21                                                            d) 11

29 tháng 5 2021

a)  a) Biến đổi vế trái thành 326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (6x+136x+6x):6x và làm tiếp

20 tháng 8 2021

`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`

`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`

`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`

`=((9sqrt6)/6+(4sqrt6)/6)^2-24`

`=((13sqrt6)/6)^2-24`

`=13^2/6-24`

`=25/6`

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

11 tháng 9 2017

1. b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)

=\(\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{9}}+\sqrt{6x}\right):\sqrt{6x}\)

=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)

=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}\)

2.

P=\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)(bn có ghi sai đề ko)

a) ĐKXĐ : \(x\ge1,x\ge2,x\ge0\)

b) P=\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+4\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

c) thay x= \(4-2\sqrt{3}\)vào P ta có :

\(\dfrac{1}{\sqrt{4-2\sqrt{3}}-2}=\dfrac{1}{\sqrt{3}-1-2}=\dfrac{1}{\sqrt{3}-3}\)

13 tháng 9 2017

@Lê Đình Thái mk k ghi sai dè nha bn

AH
Akai Haruma
Giáo viên
17 tháng 7 2021

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

AH
Akai Haruma
Giáo viên
17 tháng 7 2021

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

8 tháng 5 2022

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(\left(\dfrac{6x+4\sqrt{x}}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}+6}+1\right):\dfrac{1}{\sqrt{x}-2}\)

\(=\left(\dfrac{2\sqrt{x}\left(3\sqrt{x}+2\right)}{\sqrt{x}\left(x-4\right)}-\dfrac{6}{3\left(\sqrt{x}+2\right)}+1\right):\dfrac{1}{\sqrt{x}-2}\)

\(=\left(\dfrac{2\left(3\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}+2}+1\right):\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{6\sqrt{x}+4-2\left(\sqrt{x}-2\right)+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{1}\)

\(=\dfrac{x+6\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)}\)

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