M=\(\frac{x^2-9}{5x-10}:\frac{x^2+3x}{x-2}\)
a)ĐKXĐ
b)Rút gọn
c)tính M khi x=1/2
d)tính x khi M=1/2
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14 tháng 6 2018
M = ( x - 5)( x + 2 ) + ( 3x - 6 )( x + 2 ) - ( 3x - 1/2 )2 + 5x2
= x2-3x-10+3x2-12-(9x2-3x+1/4)+5x2
= x2-3x-10+3x2-12-9x2+3x-1/4+5x2
= 0.x - 89/4
Thay x=2018 => M= -89/4
HH
14 tháng 6 2018
\(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)
\(M=x^2+2x-5x-10+\left(3x^2+6x-6x-12\right)-\left(9x^2-\frac{3}{2}x+\frac{1}{4}\right)+5x^2\)
\(M=x^2-3x-10+3x^2-12-9x^2+\frac{3}{2}x-\frac{1}{4}+5x^2\)
\(M=-\frac{3}{2}x-\frac{41}{4}\)
Thay x = 2018 vào biểu thức \(M=-\frac{3}{2}x-\frac{41}{4}\), ta có:
\(M=-\frac{3}{2}.2018-\frac{41}{4}=-3027-\frac{41}{4}=\frac{-12149}{4}\)
Vậy giá trị của biểu thức \(M=\left(x-5\right)\left(x+2\right)+\left(3x-6\right)\left(x+2\right)-\left(3x-\frac{1}{2}\right)^2+5x^2\)khi x = 2018 là \(-\frac{12149}{4}\)
25 tháng 1 2018
a)\(\text{ĐKXĐ:}\hept{\begin{cases}x^3-4x\ne0\\6-3x\ne0\\x+2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne\mp2\end{cases}}\)
\(M=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right].\frac{x+2}{6}\)
\(=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)
\(=\frac{1}{x+2}\)
b) /x/= \(\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
*\(\text{Với }x=\frac{1}{2}\text{ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{\frac{1}{2}+2}=\frac{2}{5}\)
*\(\text{Với x= -1/2 ta có pt:}\)
\(M=\frac{1}{x+2}=\frac{1}{-\frac{1}{2}+2}=\frac{2}{3}\)
27 tháng 1 2018
a) = (\(\frac{x^2}{x\left(x^2\right)-4}+\frac{6}{3\left(2-x\right)}+\frac{1}{x+2}\)):(x-2+\(\frac{10-x^2}{x+2}\))
=(\(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}+\frac{-6}{3\left(x-2\right)}+\frac{1}{x+2}\)) :(x-2+\(\frac{10-x^2}{x+2}\))
=(\(\frac{3x^2-6x\left(x+2\right)+\left(x-2\right)3x}{3x\left(x-2\right)\left(x+2\right)}\)) :(\(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\))
=(\(\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}\)):(\(\frac{x^2-4+10-x^2}{x+2}\))
=\(\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\):\(\frac{6}{x+2}\)
=\(\frac{-6}{\left(x-2\right)\left(x+2\right)}\):\(\frac{6}{x+2}\)
=\(\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
=\(\frac{-1}{x-2}\)
Vậy M=\(\frac{-1}{x-2}\)
b)Vì /x/ =1/2 nên x=1/2 hoặc x=-1/2Thay x=1/2 vào M ta được;
\(\frac{-1}{\frac{1}{2}-2}\)=\(\frac{2}{3}\)
Thay x=-1/2 vào M ta được:
\(\frac{-1}{-\frac{1}{2}-2}\)=\(\frac{2}{5}\)
Vậy \(M\in\)\(\hept{\begin{cases}\\\end{cases}\frac{2}{5};\frac{2}{3}}\)khi /x/=1/2
15 tháng 12 2017
\(M=\left(\frac{x^2+x+10}{x^2-9}-\frac{1}{x-3}\right):\frac{1}{x-3}\) \(ĐKXĐ:x\ne\pm3\)
\(M=\left[\frac{x^2+x+10}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right]:\frac{1}{x-3}\)
\(M=\left[\frac{x^2+x+10-x-3}{\left(x-3\right)\left(x+3\right)}\right]:\frac{1}{x-3}\)
\(M=\frac{x^2+7}{\left(x-3\right)\left(x+3\right)}.\left(x-3\right)\)
\(M=\frac{x^2+7}{x+3}\)
vậy \(M=\frac{x^2+7}{x+3}\)
khi \(x=5\)thì \(M=\frac{5^2+7}{5+3}=\frac{25+7}{8}=\frac{32}{8}=4\)
vậy \(M=4\)khi \(x=5\)
24 tháng 11 2019
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
15 tháng 12 2018
\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
15 tháng 12 2018
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem
NN
17 tháng 10 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)
\(M=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)
\(=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\left[\frac{x^2+3x+2}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x^2+9x}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}.\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2-8x^2}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-4x^2\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-2x\right)\left(1+2x\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x}{3x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x+x^2-3x-1}{3x}=\frac{x^2-x}{3x}=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b) Với \(x=6013\)( thỏa mãn ĐKXĐ )
Thay \(x=6013\)vào biểu thức ta được:
\(M=\frac{6013-1}{3}=\frac{6012}{3}=2004\)
\(M=\frac{x^2-9}{5x-10}:\frac{x^2+3x}{x-2}\)
\(M=\frac{\left(x-3\right)\left(x+3\right)}{5\left(x-2\right)}:\frac{x\left(x+3\right)}{x-2}\)
\(M=\frac{\left(x-3\right)\left(x+3\right)}{5\left(x-2\right)}.\frac{x-2}{x\left(x+3\right)}\)
không bảo rút gọn nhưng mình vẫn rút gọn cho dễ làm nhé :)))
\(M=\frac{\left(x-3\right)\left(x+3\right)\left(x-2\right)}{5\left(x-2\right).x\left(x+3\right)}=\frac{x-3}{5x}\) (1)
a) ĐKXĐ: \(x\ne0;x\ne2;x\ne-3\)
b) thay x = 1/2 vào (1), ta có: \(M=\frac{\frac{1}{2}+3}{5.\frac{1}{2}}=\frac{7}{5}\)
c) \(\frac{x-3}{5x}=\frac{1}{2}\)
<=> 2(x - 3) = 5x
<=> 2x - 6 = 5x
<=> 2x - 6 - 5x = 0
<=> -3x - 6 = 0
<=> -3x = 0 + 6
<=> -3x = 6
<=> x = -2