9+x=9+9.Tìm x
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280 - x.9 = 450
x.9 = 280 - 450
x.9 = -170
x= -170/9
`5/9+4/9:x=1/3`
`=>4/9:x=1/3-5/9`
`=>4/9:x=3/9-5/9`
`=>4/9:x=-2/9`
`=>x=4/9:(-2/9)`
`=>x=4/9.(-9/2)`
`=>x=-4/2`
`=>x=-2`
b, \(B=\frac{\frac{x}{x+3}-\frac{9}{x^2+6x+9}}{\frac{3}{x+3}}=\frac{\frac{x}{x+3}-\frac{3^2}{x^2+2\cdot3\cdot x+3^2}}{\frac{3}{x+3}}\)
\(=\frac{\frac{x}{x+3}-\left(\frac{3}{x+3}\right)^2}{\frac{3}{x+3}}=1-\frac{3}{x+3}\)
a, Vậy điều kiện là \(x\ne3\)
c, \(B=\frac{1}{3}\Leftrightarrow1-\frac{3}{x+3}=\frac{1}{3}\)
\(\Rightarrow\frac{3}{x+3}=\frac{2}{3}\Leftrightarrow x=\frac{3}{2}\)
8367-x : 9 =4395
x:9 =8367-4395
x:9 =3972
x =3972*9
x =35748
Hok tốt^^
1.
=3/5x(3/7+4/7)+2/5x(13/9-4/9)
=3/5x1+2/5x1
=3/5+2/5
=1
2.Xx(3/4+4/5)=7/10
Xx31/20=7/10
X =7/10:31/20
X =14/31
\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)
\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)
\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)
_________________________________________
\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)
\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)
\(\frac{31}{20}\cdot x=\frac{7}{10}\)
\(x=\frac{7}{10}:\frac{31}{20}\)
\(x=\frac{14}{31}\)
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
9+x=9+9
9+x=18
x=18-9
x=9
x = 18 - 9
x = 9
Duyệt mk nhé bn !!!