2x/x+1+3*(x- 1)/x=5
giúp mik
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3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5
1/3 x (2,48+0,52) x y : 60 : 5 = 2/5
1/3 x 3 x y : 60 : 5 = 2/5
y: 60 : 5 =2/5
y: 60 =2/5 x 5
y : 60 =2
y = 2 x 60
y =120
Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)
\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)
hay y=120
a, Ta có : \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)
\(\Rightarrow-2x=-2+\dfrac{3}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{8}\)
Vậy ...
b, Ta có : \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(-\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{29}{6}\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=-\dfrac{12}{145}\)
\(\Rightarrow-\dfrac{2}{3}x=-\dfrac{12}{145}+\dfrac{3}{5}=\dfrac{15}{29}\)
\(\Rightarrow x=-\dfrac{45}{58}\)
Vậy...
x - 18 - 42 = 23 - 43 - 70 - x
\(\rightarrow\) 2x = -30
\(\rightarrow\) x = -15
\(\left(-\left(x+15\right)\right)^2\) - 19 = \(3^2\) .5
\(\rightarrow\)\(x^2+30x+225\) - 19 = 45
\(\rightarrow x^2+30x+161=0\)
\(\rightarrow x^2+23x+7x+161=0\)
\(\rightarrow x\left(x+23\right)+7\left(x+23\right)=0\)
\(\rightarrow\left(x+7\right)\left(x+23\right)=0\)
\(\rightarrow\) x = -7 hoặc x = -23
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3>=-5\\2x-3< =5\end{matrix}\right.\Leftrightarrow-1< =x< =4\)
a) \(\dfrac{3}{5}\times x=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}\div\dfrac{3}{5}\)
\(x=\dfrac{20}{21}\)
b) \(\dfrac{1}{8}\div x=\dfrac{1}{5}\)
\(x=\dfrac{1}{8}\div\dfrac{1}{5}\)
\(x=\dfrac{5}{8}\)
a: \(x\cdot12=621:5\)
=>\(x\cdot12=124,2\)
=>\(x=\dfrac{124.2}{12}=10.35\)
b: \(2.7\cdot x< 5\)
=>\(x< \dfrac{5}{2,7}\)
=>\(x< \dfrac{50}{27}\)
ĐK: \(x\ne-1;x\ne0\)
\(\Leftrightarrow\frac{2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)}{x\left(x+1\right)}=0\)
\(\Rightarrow2x^2+3\left(x^2-1\right)-5x^2-5x=0\)
\(\Leftrightarrow5x^2-3-5x^2-5x=0\)
\(\Leftrightarrow x=-\frac{3}{5}\left(\text{thỏa mãn ĐK}\right)\)