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x2−10x+25x2−5x=(x−5)2x(x−5)=x−5x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">x2−10x+25x2−5x=(x−5)2x(x−5)=x−5x

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25" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">25

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c) tự làm, đkxđ: x1;x1

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nguyễn hải đăng
19 tháng 12 2019 lúc 21:50

ê k bn với mk ik

😘 😘 😘 😘

Thùy Nguyễn
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Nguyễn Phương HÀ
29 tháng 6 2016 lúc 22:42

Toán lớp 8

Shinchan-XYZ
30 tháng 6 2016 lúc 6:43

chưa họclolang

Nguyễn Thị Khánh Ly
17 tháng 12 2017 lúc 10:57

DADYthanghoa

Karry Nhi
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Minh Triều
13 tháng 1 2016 lúc 13:45

ĐKXĐ : x2-5x khác 0

<=>x.(x-5) khác 0

<=> x khác 0 và x khác 5

a)

\(\frac{x^2-10x+25}{x^2-5x}=0\Rightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\)

<=>x-5=0

<=>x=5

Mà x khác 5 nên không có x nào thỏa mãn phân thức bằng 0

b)\(\frac{x^2-10x+25}{x^2-5x}=\frac{5}{2}\Leftrightarrow\frac{\left(x-5\right)^2}{x.\left(x-5\right)}=\frac{5}{2}\Leftrightarrow\frac{x-5}{x}=\frac{5}{2}\Leftrightarrow\frac{2.\left(x-5\right)}{2x}=\frac{5x}{2x}\)

\(\Rightarrow2\left(x-5\right)=5x\Leftrightarrow2x-10=5x\Leftrightarrow-3x=10\Leftrightarrow x=-\frac{10}{3}\)

c) \(\frac{x^2-10x+25}{x^2-5x}=\frac{\left(x-5\right)^2}{x.\left(x-5\right)}=\frac{x-5}{x}=1-\frac{5}{x}\)

Để phân thức trên nguyên thì : 1-5/x là số nguyên

=>5/x là số nguyên

=>x thuộc Ư(5)={1;-1;5;-5}

Mà x khác 5 nên: x={1;-1;-5}

Vậy x={1;-1;-5}

Nguyễn Mai Phương
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Phùng Minh Quân
28 tháng 1 2018 lúc 10:35

a) Ta có  \(\left|1-x\right|\ge0\)

Dấu "=" xảy ra khi \(x=1\)và khi đó A đạt gấ trị nhỏ nhất

b) Ta có 
\(x+5=x+3+2\)chia hết cho \(x+3\)\(\Rightarrow\)\(2\)chia hết cho \(x+3\)\(\Rightarrow\)\(\left(x+3\right)\inƯ\left(2\right)\)

\(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)

Do đó :

\(x+3=1\Rightarrow x=1-3=-2\)

\(x+3=-1\Rightarrow x=-1-3=-4\)

\(x+3=2\Rightarrow x=2-3=-1\)

\(x+3=-2\Rightarrow x=-2-3=-5\)

Vậy \(x=\left\{-2;-4;-1;-5\right\}\)

Chúc bạn học tốt 

Trần Huỳnh Như
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Nguyễn Lê Phước Thịnh
30 tháng 1 2022 lúc 1:08

a; \(A=\left(\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}\right):\left(1-\dfrac{2x}{x^2+1}\right)\)

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

b: Để A<0 thì x-1<0

hay x<1

c: Để A nguyên thì \(x-1\in\left\{1;-1\right\}\)

hay \(x\in\left\{2;0\right\}\)

Trang Lê
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A Nguyễn
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Nhan Tran
16 tháng 2 2022 lúc 19:54

Ai 2k9 ko

huy ngo
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missing you =
25 tháng 12 2021 lúc 17:03

\(\left(đk:x\ne5;x\ne0\right)A=\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}=1-\dfrac{5}{x}\in Z\Leftrightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Trần Anh Tuấn
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