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Dan Lin
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 Mashiro Shiina
30 tháng 1 2018 lúc 13:34

\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)

\(VP=2013+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)

\(VP=1+\left(\dfrac{2012}{2}+1\right)+....+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{1}{2013}+1\right)\)

\(VP=\dfrac{2014}{2014}+\dfrac{2014}{2}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}\)

\(VP=2014\left(\dfrac{1}{2}+..+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)

\(VP-VT=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)-x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)=0\)

\(\Rightarrow\left(2014-x\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)=0\)

\(\Rightarrow x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\ne0\right)\)

Cao Ngọc Minh Hân
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Incursion_03
2 tháng 1 2019 lúc 22:25

Đề sai kìa ... sao mẫu số lại bằng 0 ?

Vũ Nhược Ann
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Vũ Nhược Ann
7 tháng 2 2020 lúc 11:01

a, 8/x-8 + 11/x-11 = 9/x-9  + 10/ x-10

b, x/x-3 - x/x-5 = x/x-4 - x/x-6

c, 4/x^2-3x+2  - 3/2x^2-6x+1   +1 = 0

d, 1/x-1 + 2/ x-2  + 3/x-3  = 6/x-6

e, 2/2x+1 - 3/2x-1 = 4/4x^2-1

f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3

g, 1/x-1 + 2x^2 -5/x^3 -1  = 4/ x^2 +x+1

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nguyễn thành tâm
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Iruky Hita
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NTB OFFICIAL
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Akai Haruma
4 tháng 7 2018 lúc 15:37

1)

ĐK: \(x,y\neq 0\); \(x+y\neq 0\)

\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)

\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)

2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)

\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)

\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)

Akai Haruma
4 tháng 7 2018 lúc 15:46

3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)

\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)

4) ĐK: \(x\neq \frac{\pm 1}{3}\)

\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)

\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)

\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)

5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)

\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)

\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)

\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)

\(=\frac{3}{(x+1)^2}\)

Akai Haruma
4 tháng 7 2018 lúc 15:57

6) ĐK: \(x\neq \pm 3\); 0

\(\left(\frac{3}{x-3}+\frac{2x}{x^2-9}+\frac{x}{x+3}\right): \frac{2x}{x+3}\)

\(=\left(\frac{3(x+3)}{(x-3)(x+3)}+\frac{2x}{(x-3)(x+3)}+\frac{x(x-3)}{(x+3)(x-3)}\right). \frac{x+3}{2x}\)

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

\(\frac{(x^2+2x+9)(x+3)}{(x-3)(x+3).2x}=\frac{x^2+2x+9}{2x(x-3)}\)

7) ĐK: \(x\neq 2; \pm 3;0\)

\(\left(\frac{3}{x^2-9}+\frac{1}{x^2+3x}-\frac{1}{x^2-3x}\right): \frac{x-2}{2x^2+6x}\)

\(=\left(\frac{3x}{x(x-3)(x+3)}+\frac{x-3}{x(x-3)(x+3)}-\frac{x+3}{(x+3)x(x-3)}\right).\frac{2x(x+3)}{x-2}\)

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

\(=\frac{3x-6}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}=\frac{3(x-2).2x(x+3)}{x(x-3)(x+3)(x-2)}=\frac{6}{x-3}\)

ytr
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Nguyễn Việt Lâm
14 tháng 11 2019 lúc 12:21

ĐKXĐ; ...

a/ \(P=\frac{x^2}{x+4}\left[\frac{\left(x+4\right)^2}{x}\right]+9=x\left(x+4\right)+9=\left(x+2\right)^2+5\ge5\)

\(P_{min}=5\) khi \(x=-2\)

b/ \(Q=\left(\frac{\left(x+2\right)\left(x^2-2x+4\right).4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right).\frac{x\left(x-2\right)^3}{-16}\)

\(=\left(\frac{4\left(x^2-2x+4\right)-4x\left(x-2\right)}{\left(x-2\right)^2}\right).\frac{-x\left(x-2\right)^3}{16}\)

\(=\frac{16}{\left(x-2\right)^2}.\frac{-x\left(x-2\right)^3}{16}=-x\left(x-2\right)=-x^2+2x\)

\(=1-\left(x-1\right)^2\le1\)

\(Q_{max}=1\) khi \(x=1\)

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NGUYỄN THỊ THU HOÀI
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nguyễn ngọc quỳnh anh
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