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\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
14 . 16 . 25 = (14 . 25) . 16
= 400 . 16
= 4 . 16 . 100
= 64 . 100
= 6400
a) \(\dfrac{3}{8}+\dfrac{15}{-25}+\dfrac{3}{5}\)
\(=\dfrac{-9}{40}+\dfrac{3}{5}\)
\(=\dfrac{3}{8}\)
b) \(\dfrac{-5}{18}+\dfrac{23}{45}-\dfrac{9}{10}\)
\(=\dfrac{7}{30}-\dfrac{9}{10}\)
\(=\dfrac{-2}{3}\)
c) \(\dfrac{-5}{12}+\dfrac{15}{18}-2,25\)
\(=\dfrac{5}{12}-2,25\)
\(=\dfrac{-11}{6}\)
d) \(\dfrac{5}{6}+\dfrac{2}{3}-0,5\)
\(=\dfrac{3}{2}-0,5\)
\(=1\)
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
a) a mũ 3 x a mũ 9 là : a3 x b9
b) ( a mũ 5 ) mũ 7 là : (a5)7
c) ( 2 mũ 3 ) mũ 5 x ( 2 mũ 3 ) mũ 3 là : (23)5 x (23)3
Hok tốt !
Đầu bài là thế ạ còn viết thì em ko biết anh giải giùm em ạ
a) \(\dfrac{-5}{11}+\left(\dfrac{-6}{11}+1\right)\)
\(=\dfrac{-5}{11}+\left(\dfrac{-6}{11}+\dfrac{11}{11}\right)\)
\(=\dfrac{-5}{11}+\dfrac{5}{11}\)
\(=0\)
b) \(\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{-2}{3}+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\)
c) \(\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)
\(=\dfrac{-1}{4}+\dfrac{-3}{8}+\dfrac{5}{8}\)
\(=\dfrac{-2}{8}+\dfrac{-3}{8}+\dfrac{5}{8}\)
\(=0\)
d) \(\dfrac{3}{4}.\dfrac{7}{25}+\dfrac{3}{4}.\dfrac{18}{25}\)
\(=\dfrac{3}{4}.\left(\dfrac{7}{25}+\dfrac{18}{25}\right)\)
\(=\dfrac{3}{4}.1\)
\(=\dfrac{3}{4}\)
Chúc bạn học tốt
`a)25/(x+1)-1 1/6=-1/3-0,5`
`=>25/(x+1)=-1/3-1/2+1+1/6`
`=>25/(x+1)=1/3`
`=>75=x+1`
`=>x=74`
Vậy `x=74`
`b)(2x+25 3/5)^2-9/25=0`
`=>(2x+128/5)=9/25`
`**2x+128/5=3/5`
`=>2x=-125/5=-25`
`=>x=-25/2`
`**2x+128/5=-3/5`
`=>2x=-131/5`
`=>x=-131/10`
cám ơn bạn nhiều lắm ạ