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Sách Giáo Khoa
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Quốc Đạt
22 tháng 4 2017 lúc 20:56

a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0

b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)

Lưu Hạ Vy
22 tháng 4 2017 lúc 20:47

Để học tốt Toán 9 | Giải bài tập Toán 9

See you again
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MA
30 tháng 6 2019 lúc 21:36

\(M=\sqrt{\frac{m}{1-2x+x^2}}\times\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m\times\left(1-2x+x^2\right)}}{\sqrt{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m}\times\sqrt{1-2x+x^2}}{9}\)

\(=\frac{\sqrt{m}\times\sqrt{4m}}{9}\)

\(=\frac{2m}{9}\)

vậy . . .

Huy Hoang
18 tháng 11 2020 lúc 20:34

\(M=\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}.\frac{4m\left(1-x\right)^2}{81}}\)

     \(=\frac{\sqrt{4m^2}}{81}\)

     \(=\frac{\sqrt{4m^2}}{\sqrt{81}}=\frac{2m}{9}\)

Vậy : \(M=\frac{2m}{9}\) 

Khách vãng lai đã xóa
Nobi Nobita
18 tháng 11 2020 lúc 20:36

\(M=\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\sqrt{\frac{m}{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-x^2\right)}{81}}\)

\(=\frac{\sqrt{m}}{\left|1-x\right|}.\frac{\sqrt{4m\left(1-x\right)^2}}{\sqrt{81}}\)

\(=\frac{\sqrt{m}}{\left|1-x\right|}.\frac{2\sqrt{m}.\left|1-x\right|}{9}=\frac{2m}{9}\)

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Chiều Xuân
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Despacito
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Trà My
2 tháng 10 2017 lúc 21:43

\(\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(x-1\right)^2}{81}}=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{2\sqrt{m}.\left|x-1\right|}{9}=\frac{2m}{9}\)

\(x\ne1\) chứ không phải x>1 nên không thể ghi |x-1|=x-1 nhé Despacito

Despacito
2 tháng 10 2017 lúc 15:22

A..mk vua nghi ra bai nay

\(\sqrt{\frac{m}{x^2-2x+1}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{\sqrt{4m\left(x-1\right)^2}}{9}\) ( Thoa man DKXD \(m>0;x\ne1\)

\(=\frac{\sqrt{m}}{x-1}.\frac{2\left(x-1\right).\sqrt{m}}{9}\)

\(=\frac{2m}{9}\)

ko biet co dung ko nua 

nguyen thi nhu y
2 tháng 10 2017 lúc 19:12

Mình làm cách khác mà cũng ra giống kq zậy đó

Dương Thùy
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Ahwi
19 tháng 6 2019 lúc 22:13

1/ \(\sqrt{\frac{m}{1-2x+x^2}}\cdot\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-x\right)^2}{81}}\)

\(=\sqrt{\frac{m}{\left(1-x\right)^2}\cdot\frac{4m\left(1-x\right)^2}{81}}\)

\(=\sqrt{\frac{4m^2}{81}}=\sqrt{\frac{\left(2m\right)^2}{9^2}}=\frac{2\left|m\right|}{9}\)

3/\(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)

\(=\frac{a+b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)

\(=\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a+b\right|}\)

TH1: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{-\left(a+b\right)}=-\left|a\right|\)

TH2: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)

Ahwi
19 tháng 6 2019 lúc 22:27

2/\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}-a}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1}\cdot\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)

\(=\frac{\left(1-a\sqrt{a}+\sqrt{a}-a\right)\cdot\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)

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

\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)

\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}=\frac{-\left(1-a\right)^2}{\left(1-a\right)^2}=-1\)

Trang Nguyễn
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Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 11:34

e) Ta có: \(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

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

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

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

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

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

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

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

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

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

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

 

Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 11:36

m) Ta có: \(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)\)

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

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)

\(=\left(\sqrt{a}-1\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)

Nguyễn Ngọc Lộc
28 tháng 6 2021 lúc 11:37

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

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

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

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

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

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

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

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

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

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

 

 

2012 SANG
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Giáp Thị Hiền Lương
27 tháng 9 2023 lúc 3:53

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Nguyễn Quỳnh Chi
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Nguyễn Lê Phước Thịnh
31 tháng 7 2022 lúc 11:54

Bài 1:

a: ĐKXĐ: 2x+3>=0 và x-3>0

=>x>3

b: ĐKXĐ:(2x+3)/(x-3)>=0

=>x>3 hoặc x<-3/2

c: ĐKXĐ: x+2<0

hay x<-2

d: ĐKXĐ: -x>=0 và x+3<>0

=>x<=0 và x<>-3

Võ Thanh Tùng
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Nguyễn Lê Phước Thịnh
6 tháng 8 2022 lúc 14:11

Câu 1: 

a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

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

b: Để Q>0 thì \(\sqrt{a}-2>0\)

=>a>4