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a) Ta có: \(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\)
\(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}\)
\(=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}\)
\(=\dfrac{25}{36}\)
b) Ta có: \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(=\left(\dfrac{8}{10}+\dfrac{5}{10}\right):\dfrac{-5}{13}\)
\(=\dfrac{13}{10}\cdot\dfrac{13}{-5}\)
\(=-\dfrac{169}{50}\)
c) Ta có: \(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{132}\cdot\dfrac{132}{103}=\dfrac{115}{103}\)
Số tiền Nam mua sách: \(320000\times\dfrac{1}{4}=80000\) (đồng)
Số tiền Nam mua vở: \(90000:\dfrac{2}{3}=135000\) (đồng)
Số tiền Nam mua dụng cụ học tập: \(320000-\left(80000+135000\right)=105000\) (đồng)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
\(\frac{7}{2\cdot7}+\frac{7}{7\cdot12}+\frac{7}{12\cdot17}+...+\frac{7}{102\cdot107}\)
\(=\frac{7}{5}\left(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{102\cdot107}\right)\)
\(=\frac{7}{5}\cdot\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+..+\frac{1}{102}-\frac{1}{107}\right)\)
\(=\frac{7}{5}\left(\frac{1}{2}-\frac{1}{107}\right)\)
Bạn tính tiếp nhé
a) \(7^2-7\left(13-x\right)=14\)
\(7\left(13-x\right)=49-14=35\)
\(13-x=5\)
\(x=13-5=8\)
b) \(5x-5^2=10\)
\(5x=10+25=35\)
\(x=7\)
c) \(4\left(x-5\right)-2^3=2^4.3=48\)
\(4\left(x-5\right)=48+8=56\)
\(x-5=14\)
\(x=19\)
4:
a: \(\Leftrightarrow49+7\left(x-13\right)=14\)
=>7(x-13)=35
=>x-13=5
=>x=18
b: \(5x-5^2=10\)
=>\(5x=10+25=35\)
=>x=7
c: \(4\left(x-5\right)-2^3=2^4\cdot3\)
=>\(4\left(x-5\right)=16\cdot3+8=56\)
=>x-5=14
=>x=19