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TD
1
6 tháng 8 2017
\(\dfrac{2x-1}{\left(x-2\right)^2}+\dfrac{5x}{x-2}-\dfrac{25x}{5\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{\left(2x-1\right).5}{\left(x-2\right)^2.5}+\dfrac{5x\left(x-2\right).5}{\left(x-2\right).\left(x-2\right).5}-\dfrac{25x\left(x-2\right)}{5\left(x-2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{10x-5+25x^2-50x-25x^2+50x}{5\left(x-2\right)^2}=0\)
\(\Leftrightarrow\dfrac{10x-5}{5\left(x-2\right)^2}=0\)
\(\Leftrightarrow\dfrac{5\left(2x-1\right)}{5\left(x-2\right)^2}=0\)
\(\Leftrightarrow\dfrac{2x-1}{x-2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
8 tháng 1 2018
\(2x^4-5x^3-27x^2+25x+50=0\)
\(\Leftrightarrow2x^4-4x^3-x^3+2x^2-25x^2+50x+25x^2-25x+50=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-x^2\left(x-2\right)-25x\left(x-5\right)+25\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-x^2-25x+25\right)=0\)
:D sorry mk ko bt phân tích 2x^3-x^2-25x+25 :D
NN
1
26 tháng 10 2017
Ta có : M\(^2\)= (\(\dfrac{5x-4y}{5x+4y}\))\(^2\) = \(\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)= \(\dfrac{25x^2+16y^2-40xy}{25x^2+16y^2+40xy}\)
= \(\dfrac{41xy-40xy}{41xy+40xy}=\dfrac{xy}{81xy}=\dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)
Mà 4y < 5x < 0 \(\Rightarrow\)5x - 4y > 0 . 5x +4y < 0 \(\Rightarrow\) M < 0
Vậy M = - \(\dfrac{1}{9}\)
BT
0
TT
30 tháng 11 2018
a, \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)
\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)
\(=\dfrac{5x-35}{5x\left(x-7\right)}\)
\(=\dfrac{5\left(x-7\right)}{5x\left(x-7\right)}=\dfrac{1}{x}\)
b, \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)
\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x}{x\left(x-5x\right)\left(1+5x\right)}+\dfrac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)}\)\(=\dfrac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}=\dfrac{\left(5x-1\right)^2}{x.\left(1-5x\right)\left(1+5x\right)}\)
\(=\dfrac{\left(5x-1\right)^2}{-x\left(5x-1\right)\left(1+5x\right)}\) \(=\dfrac{-\left(5x-1\right)}{x\left(1+5x\right)}\)
DN
2
BN
1
27 tháng 5 2022
\(=\dfrac{y}{x\left(y-5x\right)}+\dfrac{25x-15y}{\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y\left(y+5x\right)+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+5xy+25xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
\(=\dfrac{y^2+30xy-15y^2}{x\left(y-5x\right)\left(y+5x\right)}\)
BN
1
NL
1
27 tháng 11 2018
\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y+5x\right)\cdot\left(y-5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y+5x\right)\left(y-5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}=\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{xy+5x^2}\)
\(5x^2+25x-750=0\)
\(\Leftrightarrow5\left(x^2+5x-150\right)=0\)
\(\Leftrightarrow5\left(x^2+15x-10x-150\right)=0\)
\(\Leftrightarrow5\left[\left(x^2+15x\right)-\left(10x+150\right)\right]=0\)
\(\Leftrightarrow5\left[x\left(x+15\right)-10\left(x+15\right)\right]=0\)
\(\Leftrightarrow5\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-10=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-15\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{10;-15\right\}\)