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BC
1
14 tháng 12 2020
ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)
Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)
KT
1
22 tháng 2 2021
1) Ta có: \(\dfrac{x\left|x-2\right|}{x^2-5x+6}\)
\(=\left[{}\begin{matrix}\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x< 2\right)\\\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x>2\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}\dfrac{-x}{x-3}\\\dfrac{x}{x-3}\end{matrix}\right.\)
2) Ta có: \(\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)
\(=\dfrac{\left(a^x\right)^2-\left(b^x\right)^2}{a^x-b^x}\)
\(=\dfrac{\left(a^x-b^x\right)\left(a^x+b^x\right)}{a^x-b^x}=a^x+b^x\)
18 tháng 4 2021
a, ĐK : \(x\ne1;2;3;4;5\)
b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)
16 tháng 5 2022
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}=\dfrac{x-x+5}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
12 tháng 1 2023
a: ĐKXĐ: x<>1; x<>2; x<>3
\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)
\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2x^3-4x^2}{\left(x-2\right)}\cdot\dfrac{1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\dfrac{2x^2}{x^4+x^2+1}\)
b:
28 tháng 7 2021
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0
MH
15 tháng 9 2023
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
\(=\dfrac{\left(4-x\right)\left(4+x\right)}{\left(4-x\right)^2}=\dfrac{4+x}{4-x}\)
ai chơi sửa đề vậy bạn??
\(=\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}=\dfrac{x}{x+2}\)