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\(\dfrac{-4}{13}\cdot\dfrac{5}{17}+\dfrac{-12}{13}\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{-4}{17}\cdot\dfrac{5}{17}+\dfrac{-4}{13}\cdot3\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{-5}{17}+\dfrac{-12}{17}+1\right)\)
\(=\dfrac{4}{13}\left(-1+1\right)=\dfrac{4}{13}\cdot0=0\)
Bài này chỉ cần tính ngược từ dưới lên trên nhé ^^
=\(\dfrac{1}{1+\dfrac{1}{15}}+\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{\dfrac{16}{15}}+\dfrac{1}{\dfrac{3}{2}}=\dfrac{15}{16}+\dfrac{2}{3}=\dfrac{77}{48}\)
Giải:
\(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
\(=\left(\dfrac{-3}{7}+\dfrac{4}{11}\right).\dfrac{11}{7}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right).\dfrac{11}{7}\)
\(=\left[\dfrac{-3}{7}+\dfrac{4}{11}+\left(-\dfrac{4}{7}\right)+\dfrac{7}{11}\right].\dfrac{11}{7}\)
\(=\left[\dfrac{-3}{7}+\left(-\dfrac{4}{7}\right)+\dfrac{4}{11}+\dfrac{7}{11}\right].\dfrac{11}{7}\)
\(=\left[\dfrac{-7}{7}+\dfrac{11}{11}\right].\dfrac{11}{7}\)
\(=\left[-1+1\right].\dfrac{11}{7}\)
\(=0.\dfrac{11}{7}=0\)
Vậy ...
\(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(\dfrac{-4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
= \(\left(\dfrac{-3}{7}+\dfrac{4}{11}\right).\dfrac{11}{7}+\left(\dfrac{-4}{7}+\dfrac{7}{11}\right).\dfrac{11}{7}\)
= \(\left(\dfrac{-3}{7}+\dfrac{4}{11}+\dfrac{-4}{7}+\dfrac{7}{11}\right)\)\(.\dfrac{11}{7}\)
= \(\left(-1+1\right).\dfrac{11}{7}\)
= \(0.\dfrac{11}{7}\)
= \(0\)
\(\dfrac{72-x}{7}=\dfrac{x-4}{9}\)
\(\Rightarrow9\left(72-x\right)=7\left(x-4\right)\)
\(\Rightarrow648-9x=2x-28\)
\(\Rightarrow11x-28=648\)
\(\Rightarrow11x=676\Rightarrow x=\dfrac{676}{11}\)
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x+39=259\)
\(\Rightarrow10x=220\Rightarrow x=22\)
\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=\pm10^2\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\Rightarrow x=6\\x+4=-10\Rightarrow x=-14\end{matrix}\right.\)
\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)
\(\Rightarrow x\left(x+3\right)-1\left(x+3\right)=x\left(x+2\right)-2\left(x+2\right)\)
\(\Rightarrow x^2+3x-x-3=x^2+2x-2x-4\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x-3=-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(A=\dfrac{-\left(5^2+5.3^2\right)}{5^2\left(5+3^2\right)}\)
\(A=\dfrac{-\left(5\left(5+3^2\right)\right)}{5^2\left(5+3^2\right)}\)
\(A=\dfrac{-5}{5^2}=-\dfrac{1}{5}\)
CHÚC BẠN HỌC TỐT.........
1) |x|=x+2
=> \(\left[{}\begin{matrix}x=x+2\\x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(voli\right)\\2x=-2\Rightarrow x=-1\end{matrix}\right.\)
vậy x=-1
c;b tương tự
2) \(\left|x-\dfrac{3}{2}\right|=\left|\dfrac{5}{2}-x\right|\)
=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{5}{2}-x\\x-\dfrac{3}{2}=x-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\Rightarrow x=2\\0=-1\left(voli\right)\end{matrix}\right.\)
vậy x=2
\(\frac{-464}{875}\)
\(-0,928.\frac{4}{7}=-\frac{464}{875}.\)
Chúc bạn học tốt!