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1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
1. Tìm x thuộc N:
\(\left(x-3\right)^6=\left(x-3\right)^7\)
\(\Leftrightarrow\left(x-3\right)^6-\left(x-3\right)^7=0\)
\(\Leftrightarrow\left(x-3\right)^6.\text{[}1-\left(x-3\right)\text{]}=0\)
\(\Leftrightarrow\left(x-3\right)^6.\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)(thỏa mãn \(x\in N\))
2.
Ta có: 6x=4y=3z
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
\(=\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.3=9\\z=3.4=12\end{matrix}\right.\)
1. Áp dụng TCDTSBN:
\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{a+b}{3+5}=\dfrac{32}{8}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=12\\b=20\end{matrix}\right.\)
1. Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{a+b}{3+5}=\dfrac{32}{8}=4\)
\(\dfrac{a}{3}=4\Rightarrow a=12\\ \dfrac{b}{5}=4\Rightarrow b=20\)
2. gọi độ dài 3 cạnh tam giác lần lượt là a,b,c
Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{9}\\a+b+c=630\left(m\right)\end{matrix}\right.\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{9}=\dfrac{a+b+c}{5+7+9}=\dfrac{630}{21}=30\left(m\right)\)
\(\dfrac{a}{5}=30\Rightarrow a=150\left(m\right)\\ \dfrac{b}{7}=30\Rightarrow b=210\left(m\right)\\ \dfrac{c}{9}=30\Rightarrow c=270\left(m\right)\)
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
Bài 1:
a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)
\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)
b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)
Bài 2:
Áp dụng t/c dtsbn:
\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
Lời giải :
a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=2,5\)
b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}\left(19-33\right)\)
\(=\dfrac{3}{7}\left(-14\right)\)
\(=-6\)
c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)
\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
\(=0\)
d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)
\(=-10\left(-\dfrac{7}{5}\right)\)
\(=14\)
a;\(\dfrac{1}{6}\)
câu8
a,x=1