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Bài 1
\(a,\frac{3}{5}+\left(-\frac{1}{4}\right)=\frac{7}{20}\)
\(b,\left(-\frac{5}{18}\right)\cdot\left(-\frac{9}{10}\right)=\frac{1}{4}\)
\(c,4\frac{3}{5}:\frac{2}{5}=\frac{23}{5}\cdot\frac{5}{2}=\frac{23}{2}\)
Bài 2
\(a,\frac{12}{x}=\frac{3}{4}\Rightarrow3x=12\cdot4\)
\(\Rightarrow3x=48\)
\(\Rightarrow x=16\)
\(b,x:\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^2\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^2\cdot\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^5\)
\(\Rightarrow x=-\frac{1}{243}\)
\(c,-\frac{11}{12}\cdot x+0,25=\frac{5}{6}\)
\(\Rightarrow-\frac{11}{12}x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\left(-\frac{11}{12}\right)\)
\(\Rightarrow x=-\frac{7}{11}\)
\(d,\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=-2^5\)
\(x-1=-2\)
\(x=-2+1=-1\)
Bài 3
\(\left|m\right|=-3\Rightarrow m\in\varnothing\)
Bài 3
Gọi 3 cạnh của tam giác lần lượt là a;b;c ( a,b,c>0)
Ta có
\(a+b+c=13,2\)
\(\frac{a}{3};\frac{b}{4};\frac{c}{5}\)
Ap dụng tính chất DTSBN ta có
\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{13,2}{12}=\frac{11}{10}\)
\(\hept{\begin{cases}\frac{a}{3}=\frac{11}{10}\\\frac{b}{4}=\frac{11}{10}\\\frac{c}{5}=\frac{11}{10}\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{33}{10}\\b=\frac{44}{10}=\frac{22}{5}\\c=\frac{55}{10}=\frac{11}{2}\end{cases}}\)
Vậy 3 cạnh của tam giác lần lượt là \(\frac{33}{10};\frac{22}{5};\frac{11}{2}\)
a)\(\frac{3}{5}+\left(-\frac{1}{4}\right)\)
\(=\frac{3}{5}-\frac{1}{4}\)
\(=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}\)
b)\(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)\)
\(=\frac{\left(-5\right)\left(-9\right)}{18.10}\)
\(=\frac{\left(-1\right)\left(-1\right)}{2.2}=\frac{1}{4}\)
c)\(4\frac{3}{5}:\frac{2}{5}\)
\(=\frac{23}{5}:\frac{2}{5}\)
\(=\frac{23}{5}.\frac{5}{2}\)
\(=\frac{23.1}{1.2}=\frac{23}{2}\)
1/
a)\(\frac{12}{x}=\frac{3}{4}\)
\(\Rightarrow x.3=12.4\)
\(\Rightarrow x.3=48\)
\(\Rightarrow x=48:3=16\)
b)\(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)
\(x=\left(\frac{-1}{3}\right)^2.\left(\frac{-1}{3}\right)^3\)
\(x=\frac{\left(-1\right)^2}{3^2}.\frac{\left(-1\right)^3}{3^3}\)
\(x=\frac{1}{9}.\frac{-1}{27}=-\frac{1}{243}\)
a) \(\frac{2}{9}+\frac{1}{5}+\frac{7}{9}+\frac{4}{5}=\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)= 1+1=2 \)
b) \(\frac{19}{37}+\left(1+\frac{19}{37}\right)=\left(\frac{19}{37}-\frac{19}{37}\right)+1=0+1=1\)
c) \(\frac{9}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)+\frac{7}{8}=\frac{9}{8}+\frac{7}{8}-\left(\frac{17}{7}-\frac{3}{7}\right)=2-2=0\)
d) \(\frac{3}{5}\times\frac{8}{27}\times\frac{5}{3}=\frac{3}{5}\times\frac{5}{3}\times\frac{8}{27}=1\times\frac{8}{27}=\frac{8}{27}\)
e) \(\frac{7}{19}\times\frac{1}{3}+\frac{7}{19}\times\frac{2}{3}=\frac{7}{19}\times\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}\times1=\frac{7}{19}\)
f) \(\frac{12}{5}\times4-4\times\frac{7}{5}=4\times\left(\frac{12}{5}-\frac{7}{5}\right)=4\times1=4\)
Dễ thôi
Ta có:
\(1+4\cdot1=5\)
Tiếp tục ta có:
\(2+5\cdot2=12\)
\(3+3\cdot6=21\)
Vậy số cần tìm là : \(8+11\cdot8=96\)
\(a,\left(\frac{3}{4}+\frac{3}{2}\right):\frac{7}{4}-\frac{3}{4}\)
\(=\left(\frac{3}{4}+\frac{6}{4}\right).\frac{4}{7}-\frac{3}{4}\)
\(=\frac{9}{4}.\frac{4}{7}-\frac{3}{4}\)
\(=\frac{9}{7}-\frac{3}{4}\)
\(=\frac{36}{28}-\frac{21}{28}\)
\(=\frac{15}{28}\)
\(b,\left(-5\right)^2.\frac{7}{45}-\left(-5\right)^2.\frac{11}{45}\)
\(=\left(-5\right)^2.\left(\frac{7}{45}-\frac{11}{45}\right)\)
\(=\left(-5\right)^2.\frac{-4}{45}\)
\(=25.\frac{-4}{45}\)
\(=\frac{-100}{45}\)
\(=\frac{20}{9}\)
#Chúc em học tốt
11 + 3 - 4 = 10 14 - 4 + 2 = 12 12 + 3 - 3 = 12
12 + 5 - 7 = 10 15 - 5 + 1 = 11 15 - 2 + 2 = 15
11 + 3 - 4 = 10 14 - 4 + 2 = 12 12 + 3 - 3 = 12
12 + 5 - 7 = 10 15 - 5 + 1 = 11 15 - 2 + 2 = 15
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