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BC
thực hiện phép tính
\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)
1
H
9 tháng 12 2018
\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)
\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)
\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x-14}{x+3}\)
\(=\left(\frac{1-2x-6+3x-9}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)
\(=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}=\frac{1}{x-3}\)
NT
Thực hiện phép tính
\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)
1
H
14 tháng 12 2018
\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)
\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)
\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)
\(=\left(\frac{1-2x-6+3x-9}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x-14}=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}\)
\(=\frac{1}{x-3}\)
YA
23 tháng 12 2019
bạn ơi tại sao lại bằng 2x +6, bạn có thể giải đáp cho mình đc ko
XO
17 tháng 1 2021
1) a) \(\frac{x}{x+1}+\frac{x^3-2x^2}{x^3+1}=\frac{x}{x+1}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3-x^2+x+x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{2x^3-3x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b) \(\frac{x+1}{2x-2}+\frac{3}{x^2-1}+\frac{x+3}{2x+2}=\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{x+3}{2\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x^2+4x+2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)
2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)
Thay x = 14 ; y = -15 vào biểu thức ta được
\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)
DV
20 tháng 12 2016
a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)
b. =\(\frac{2.\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x.\left(x+3\right)}=\frac{-2}{x^2}\)
Chắc chắn đúng, mik nhaaaaaa
TA
1
28 tháng 12 2015
\(\frac{3}{x-3}-\frac{x-6}{x^2+3x}=\frac{3}{x-3}-\frac{x-6}{x\left(x+3\right)}=\frac{3}{x-3}+\frac{x-6}{x\left(x-3\right)}=\frac{3x}{x\left(x-3\right)}+\frac{x-6}{x\left(x-3\right)}=\frac{3x+x-6}{x\left(x-3\right)}=\frac{4x-6}{x\left(x-3\right)}\)
ET
4 tháng 12 2018
a) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\\ =\dfrac{-5x-10}{8-4x}.\dfrac{4-2x}{x+2}\\ =\dfrac{-5\left(x+2\right)}{4\left(2-x\right)}.\dfrac{2\left(2-x\right)}{x+2}=\dfrac{-5}{2}\)
b)\(\dfrac{x^2-36}{2x+10}.\dfrac{3}{6-x}\\ =\dfrac{\left(x-6\right).\left(x+6\right)}{2\left(x+5\right)}.\dfrac{-3}{x-6}\\ =\dfrac{-3x-18}{2x+10}\)
25 tháng 4 2020
a) \(\frac{2x^3+x^2+x+6}{x^2-x+2}=\frac{\left(2x+3\right)\left(x^2-x+2\right)}{x^2-x+2}=2x+3\)
b) \(\frac{x}{x-3}-\frac{5x^2+27}{x^2-9}+\frac{x-9}{x+3}\)
\(=\frac{x}{x-3}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x-9}{x+3}\)
\(=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-9\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x-\left(5x^2+27\right)+x^3-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x^2-9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x}{x-3}\)
Vậy (6x3 – x2 – 14x + 3) : (2x – 3) = 3x2 + 4x – 1