Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Love
Xem chi tiết
Nguyễn Hồng Thắm
Xem chi tiết
Nguyễn Hồng Thắm
6 tháng 10 2018 lúc 11:12

Ai giải giúp mình bài 1 với bài 4 trước đi

your heart your love is...
Xem chi tiết
Nguyễn Ngọc Tú Uyên
Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 2 2022 lúc 0:08

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

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

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

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

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

Park Jimin
Xem chi tiết
Phùng Minh Quân
17 tháng 10 2018 lúc 20:14

\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

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

\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)

\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)

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

\(b)\) Ta có : \(R< -1\)

\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)

\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)

\(\Leftrightarrow\)\(4\sqrt{x}< 6\)

\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)

\(\Leftrightarrow\)\(x< \frac{9}{4}\)

Chúc bạn học tốt ~ 

LuKenz
Xem chi tiết
Huỳnh Diệu Linh
Xem chi tiết
nguyên công quyên
Xem chi tiết
Minh Nguyen
2 tháng 3 2020 lúc 18:42

Câu 3 :

\(ĐKXĐ:x>0\)

 \(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)

\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)

b) Để P = 3

\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)

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

\(\Leftrightarrow x-4\sqrt{x}+4=0\)

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

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

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

\(\Leftrightarrow x=4\)(tm)

Vậy để \(P=3\Leftrightarrow x=4\)

Khách vãng lai đã xóa
Minh Nguyen
2 tháng 3 2020 lúc 18:57

Câu 1 : Hình như sai đề !! Mik sửa :

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)

\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)

\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

b) Để A < 2

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

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

\(\Leftrightarrow2\sqrt{x}>3\)

\(\Leftrightarrow\sqrt{x}>1,5\)

\(\Leftrightarrow x>2,25\)

Vậy để \(A< 2\Leftrightarrow x>2,25\)

Khách vãng lai đã xóa
Minh Nguyen
2 tháng 3 2020 lúc 19:11

Câu 2 :

\(ĐKXĐ:\hept{\begin{cases}x\ne\frac{4}{25}\\x\ne9\\x\ne1\end{cases}}\)

\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A=\frac{\left(2-5\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

b) Mik quên mất cách chứng minh rồi :((

Chỉ biết : Dấu " = " xảy ra : \(\Leftrightarrow x=0\)

\(\)

Khách vãng lai đã xóa
nguyen le duy hung
Xem chi tiết
Không Tên
11 tháng 7 2018 lúc 20:04

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

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

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

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

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

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

\(=1-a+a=1\)