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\(a,\left(\frac{3}{4}+\frac{3}{2}\right):\frac{7}{4}-\frac{3}{4}\)
\(=\left(\frac{3}{4}+\frac{6}{4}\right).\frac{4}{7}-\frac{3}{4}\)
\(=\frac{9}{4}.\frac{4}{7}-\frac{3}{4}\)
\(=\frac{9}{7}-\frac{3}{4}\)
\(=\frac{36}{28}-\frac{21}{28}\)
\(=\frac{15}{28}\)
\(b,\left(-5\right)^2.\frac{7}{45}-\left(-5\right)^2.\frac{11}{45}\)
\(=\left(-5\right)^2.\left(\frac{7}{45}-\frac{11}{45}\right)\)
\(=\left(-5\right)^2.\frac{-4}{45}\)
\(=25.\frac{-4}{45}\)
\(=\frac{-100}{45}\)
\(=\frac{20}{9}\)
#Chúc em học tốt
a) 3/4x16/9-7/5:(-21/20)
=4/3-(-4/3)
=8/3
b)=7/3-1/3x[-3/2+(2/3+2)]
=7/3-1/3x[-3/2+8/3]
=7/3-1/3x7/6
=7/3-7/18
=35/18
c)=(20+37/4):9/4
=117/4:9/4
=13
d)=6-14/5x25/8-8/5:1/4
=6-35/4-32/5
=-11/4-32/5
=-183/20
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)\(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)
\(x=\left(\frac{-1}{3}\right)^2.\left(\frac{-1}{3}\right)^3\)
\(x=\frac{\left(-1\right)^2}{3^2}.\frac{\left(-1\right)^3}{3^3}\)
\(x=\frac{1}{9}.\frac{-1}{27}=-\frac{1}{243}\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)
\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)
a,
2/7:2/5=4/35
b,
(8/15+7/15):2/11=11/2