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\(B=\dfrac{2a+3}{a-2}=\dfrac{2\left(a-2\right)+7}{a-2}\\ =2+\dfrac{7}{a-2}\) (a nguyên, a khác 2)
Để B đạt gt nguyên thì: \(\dfrac{7}{a-2}\) cũng phải đạt gt nguyên
\(\Rightarrow7⋮\left(a-2\right)\)
\(\Rightarrow a-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\\ \Rightarrow a\in\left\{3;1;9;-5\right\}\left(TMDK\right)\)
\(x\times62+x\times48=4200\\ x\times\left(62+48\right)=4200\\ x\times110=4200\\ x=4200:110\\ x=\dfrac{420}{11}\)
X x 62 + X x 48 = 4200
X x (62 + 48) = 4200
X x 120 = 4200
X = 4200 : 120
X = 35
Vậy: ...
\(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\\ =\left(\dfrac{-3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\\ =\dfrac{0}{5}+\dfrac{1}{3}\\ =0+\dfrac{1}{3}\\ =\dfrac{1}{3}\)
Sau hai lần lấy ra, số kg gạo còn lại trong bao chiếm số phần so với bao gạo ban đầu là:
\(1-\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
Ban đầu bao gạo có số kg gạo là:
\(18:\dfrac{3}{8}=48\) (kg gạo)
Đáp số: 48kg gạo
\(\dfrac{3}{2}+\dfrac{19}{6}+\dfrac{61}{12}+...+\dfrac{463}{42}\)
\(=1+3+5+...+11+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}\right)\)
Ta đặt biểu thức: 1 + 3 + 5 + ... + 11 = A
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}\) = B
A = (11 + 1) x 6 : 2 = 36
B = Không rõ quy luật.
a) \(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{5}{4}-x=5\dfrac{1}{3}:0,8\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}\times\dfrac{5}{4}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=-\dfrac{65}{12}\)
b) \(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\times\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=\dfrac{-119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ \Rightarrow x=-\dfrac{1190}{9}\)
a)
\(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{16}{3}:\left(\dfrac{5}{4}-x\right)=\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=\dfrac{-65}{12}\)
b)
\(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\cdot\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=-\dfrac{119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ =-\dfrac{1190}{9}\)
a; (\(\dfrac{3}{4}\))4.(\(\dfrac{8}{9}\))2
= (\(\dfrac{3}{2^2}\))4.(\(\dfrac{2^3}{3^2}\))2
= \(\dfrac{3^4}{2^8}\).\(\dfrac{2^6}{3^4}\)
= \(\dfrac{3^4.2^6}{3^4.2^6}\). \(\dfrac{1}{2^2}\)
= \(\dfrac{1}{2^2}\)
= \(\dfrac{1}{4}\)
b; (\(\dfrac{-3}{5}\))6.(-\(\dfrac{5}{3}\))5
= \(\dfrac{3^6}{5^6}\).\(\dfrac{\left(-5\right)^5}{3^5}\)
= \(\dfrac{-5^5.3^5}{5^5.3^5}\).\(\dfrac{3}{5}\)
= - 1.\(\dfrac{3}{5}\)
= - \(\dfrac{3}{5}\)