Tim x , biet:\(\frac{^{x^2}}{x-2}=\frac{3x+1}{3}\)
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a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
a, 60%x + 2/3x =1/3.6 1/3
3/5x +2/3x =1/3.19/3
x.(3/5+2/3)=19/9
x.(9/15+10/15)=19/9
x.19/15=19/9
x=19/9:19/15
x=15/9
Vậy x=15/9
b,3.(3x-1/2)^3 +1/9=0
3.(3x-1/2)^3= -1/9
(3x-1/2)^3= -1/9:3
(3x-1/2)^3= -1/27
(3x-1/2)^3=(-1/3)^3
3x-1/2= -1/3
3x= -1/3-1/2
3x= -2/6+(-3/6)
3x= -5/6
x= -5/6 :3
x=-5/18
Vậy x=-5/18
a) \(P=\frac{3x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)
\(P=\frac{3\left(x-9\right)}{\left(x-3\right)\left(x-2\right)}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)
\(P=\frac{3}{x-2}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)
\(P=\frac{3\left(3-x\right)-\left(x+3\right)\left(3-x\right)-\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(3-x\right)}\)
\(P=\frac{9-3x-9+x^2-2x^2+4x-x+2}{\left(x-2\right)\left(3-x\right)}\)
\(P=\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}\) (*)
b) Thay \(x=-\frac{1}{2}\) vào (*) ta có:
\(P=\frac{2-\left(-\frac{1}{2}\right)^2}{\left[\left(-\frac{1}{2}\right)-2\right]\left[3-\left(-\frac{1}{2}\right)\right]}=\frac{2-\frac{1}{4}}{-\frac{5}{2}.\frac{7}{2}}=-\frac{\frac{7}{4}}{\frac{5}{2}.\frac{7}{2}}=-\frac{7}{35}=-\frac{1}{5}\)
c) \(\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}< 0\)
\(\Leftrightarrow2-x^2< 0\)
\(\Leftrightarrow-x^2< -2\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow\hept{\begin{cases}x< -\sqrt{2}\\-\sqrt{2}< x< \sqrt{2}\\x>2\end{cases}}\)
Vậy: ...
a, \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{3}{2}x=\frac{-5}{4}\)
\(\Leftrightarrow x=\frac{-5}{6}\)
b, \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
\(\Leftrightarrow\frac{3}{5}.\left(3x-\frac{37}{10}\right)=\frac{-57}{10}\)
\(\Leftrightarrow3x-\frac{37}{10}=\frac{-19}{2}\)
\(\Leftrightarrow3x=\frac{-29}{5}\)
\(\Leftrightarrow x=\frac{-29}{15}\)
a) \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=-\frac{3}{4}\)
\(x\left(\frac{2}{3}+\frac{5}{6}\right)+\frac{1}{2}=-\frac{3}{4}\)
\(x\cdot\frac{3}{2}=\frac{-5}{4}\)
\(x=-\frac{5}{6}\)
\(\frac{2}{5}+\frac{3}{5}\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\left(3x-\frac{37}{10}\right)=-\frac{57}{10}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=\frac{-29}{5}\)
\(x=\frac{-29}{15}\)
Bài làm:
c) \(\left(x-2\right)\left(x+3\right)>0\)
Ta xét 2 trường hợp sau:
+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)
+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)
Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)
d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{27}\)
Vậy \(x=\frac{1}{27}\)
Học tốt!!!!
a) x = 6 ; y = 15.
x = -6 ; y = -15.
b) x = 2 ; y = 2.
x = -2 ; y = -2.
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6