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a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)
\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)
b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)
a: \(-15-\left(-13+30\right)\)
\(=-15+13-30\)
\(=\left(-15-30\right)+13\)
=-45+13
=-32
b: \(225-150:\left(30+3^2\cdot5\right)\)
\(=225-150:\left(30+9\cdot5\right)\)
\(=225-150:75\)
=225-2
=223
c: \(18\cdot64+18\cdot36-1200\)
\(=18\left(64+36\right)-1200\)
\(=18\cdot100-12\cdot100\)
\(=6\cdot100=600\)
d: \(80-\left[130-\left(12-4\right)^2\right]\)
\(=80-130+\left(12-4\right)^2\)
\(=-50+8^2=64-50=14\)
a)-1-2-3-4-5-6-....-80
=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)
Khoảng cách giữa các số:(-1)-(-2)=1
Tổng các số hạng:(-1)-(-80)+1=80 số
Tổng:[(-1)+(-80)].80:2= -3240
=>-1-2-3-4-5-6+......-80=-3240
b,1-2+3-4+5-6+......+2021-2022
=(1-2)+(3-4)+(5-6)+...+(2021-2022)
=(-1)+(-1)+(-1)+...+(-1)
Tổng số cặp là:
(2022-1+1):2=1011 cặp
-1.1011=-1011
=>1-2+3-4+5-6+......+2021-2022= -1011
c, Đề bài sai
d,-4-8-12-16-.......-2020
=-4+(-8)+(-12)+(-16)+...+(-2020)
Khoảng cách giữa các số:-4-(-8)=4
Tổng các số hạng:[-4-(-2020]:4+1=505 số
Tổng:[-4+(-2020)].505:2=-511060
=>-4-8-12-16-.......-2020=-511060
Thực hiện phép tính:
a, 5/17 + -5/34 . 2/5=5/17 + -1/17=4/17
b, 1/2 . 5/6 + 2/3 . 3/4 =5/12 + 6/12 =11/12
c, ( -2/5 + 1/3 ) . ( 3/2 - 3/7 )=-1/15 . 15/14=1/14
a. \(\dfrac{5}{17}+\dfrac{-5}{34}.\dfrac{2}{5}\)
= \(\dfrac{5}{17}+\dfrac{1}{-17}\)
= \(\dfrac{5}{17}+\dfrac{-1}{17}\)
= \(\dfrac{4}{17}\)
b. \(\dfrac{1}{2}.\dfrac{5}{6}+\dfrac{2}{3}.\dfrac{3}{4}\)
= \(\dfrac{5}{12}+\dfrac{1}{2}\)
= \(\dfrac{5}{12}+\dfrac{6}{12}\)
= \(\dfrac{11}{12}\)
c. \(\left(\dfrac{-2}{5}+\dfrac{1}{3}\right).\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)
= \(\left(\dfrac{-6}{15}+\dfrac{5}{15}\right).\left(\dfrac{21}{14}-\dfrac{6}{14}\right)\)
= \(\dfrac{-1}{15}.\dfrac{15}{14}\)
= \(\dfrac{-1}{14}\)
d. \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right)\)
= \(\left(\dfrac{2}{2}+\dfrac{1}{2}\right).\left(\dfrac{3}{3}+\dfrac{1}{3}\right).\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\)
= \(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}\)
= \(\dfrac{5}{2}\)
a: \(=\dfrac{5}{34}\cdot\dfrac{2}{5}=\dfrac{2}{34}=\dfrac{1}{17}\)
b: \(=\dfrac{5}{12}+\dfrac{6}{12}=\dfrac{11}{12}\)
c: \(=\dfrac{-6+5}{15}\cdot\dfrac{21-6}{15}=-\dfrac{1}{15}\)
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
a, 80 - 4 . 5 2 - 3 . 2 3
= 80 – (4.25 – 3.8)
= 80 – 76 = 4
b, 5 6 : 5 4 + 2 3 . 2 2 - 1 2018
= 5 2 + 2 5 - 1
= 25 + 32 – 1 = 56
c, [36.4 – 4. 82 - 7 . 11 2 ]:4 – 2019 0
= (144 – 4. 5 2 ):4 – 1
= (144 – 100):4 – 1
= 11 – 1 = 10
d, 303 – 3.{[655 – (18:2+1). 4 3 +5]}: 10 0
= 303 – 3.(655 – 10.64 + 5):1
= 303 – 10 = 293