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a: \(=-4\cdot\dfrac{2}{5}=-\dfrac{8}{5}\)
b: \(=3.6\cdot\dfrac{4}{3}=4.8\)
c: \(=72000\cdot\dfrac{3}{4}=54000\left(đồng\right)\)
1/4 của 40m là:
40 : 1/4 = 160 ( m )
3/5 của 48 cm là:
48 : 3/5 = 80 ( cm )
10% của 4/5 kg là:
45 : 10 x 100= 450 ( kg )
1 và 2/13 = 15/13 ; 1 và 1/6 = 7/6
Vậy 15/13 của 7/6 m2 là:
7/6 : 15/13 = 91/90 ( m2)
Đ/s: ..........
~Hok tốt ~
a, Ta có : M= 1 - 2 + 3 -4 + 5-6 +... + 19 - 20
= ( 1-2 ) + ( 3 - 4 ) + ( 5-6 ) + ...+ ( 19 - 20 )
= ( -1 ) + ( -1 ) + ( -1 ) +...+ (-1 ) ( có 10 số ( - 1 )
= 10 . (-1 )
Bài 1:
$M=3.4.5+4.5.6+...+13.14.15$
$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$
$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$
$=-2.3.4.5+13.14.15.16=43560$
$M=43560:4=10890$
Bài 2:
a.
$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$
$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$
$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
$M=\frac{99}{100}:3=\frac{33}{100}$
1.
a) \(\frac{13}{8}:\frac{13}{4}=\frac{1}{2}=0,5=50\%\)
b) \(12,5:2,5=5=500\%\)
~ Hok tốt ~
2.
\(\left(\frac{37}{9}+\frac{13}{4}\right).\frac{9}{4}+\frac{11}{4}\)
\(=\frac{265}{36}.\frac{9}{4}+\frac{11}{4}\)
\(=\frac{265}{16}+\frac{11}{4}\)
\(=\frac{309}{16}\)
\(1+\left(\frac{9}{10}-\frac{4}{5}\right):\frac{19}{6}\)
\(=1+\frac{1}{10}:\frac{19}{6}\)
\(=1+\frac{3}{95}\)
\(=\frac{98}{95}\)
\(\left(-7+\left|13\right|\right)-\left(13-\left|-7\right|-25\right)-\left(25+\left|-10\right|-9\right)\)
\(=\left(-7+13\right)-\left(13-7-25\right)-\left(25+10-9\right)\)
\(=-7+13-13+7+25-25-10+9\)
\(=\left(-7+7\right)+\left(13-13\right)+\left(25-25\right)-10+9\)
\(=0+0+0-10+9\)
\(=-10+9\)
\(=-1\)
~ Hok tốt ~
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
a: \(=\dfrac{-3}{7}-\dfrac{4}{7}+\dfrac{5}{13}=\dfrac{5}{13}-1=-\dfrac{8}{13}\)
b: \(=\dfrac{-5}{21}-\dfrac{2}{21}+\dfrac{1}{3}=\dfrac{-7}{21}+\dfrac{1}{3}=0\)
c: \(=\dfrac{-1}{2}+\dfrac{4}{5}+\dfrac{2}{3}-\dfrac{2}{3}+\dfrac{7}{9}\)
\(=\dfrac{-45}{90}+\dfrac{72}{90}+\dfrac{70}{90}=\dfrac{97}{90}\)
Câu 2:
a: \(x^{10}=1^x\)
\(\Leftrightarrow x^{10}=1\)
=>x=1 hoặc x=-1
b: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
hay \(x\in\left\{\dfrac{15}{2};8;7\right\}\)
c: \(x^{10}=x\)
\(\Leftrightarrow x\left(x^9-1\right)=0\)
=>x=0 hoặc x=1