Thực hiện phép tính:
A) (-3) ✕ (-2)✕(-5)✕4=
B)3✕2✕(-8)✕(-5)=
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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) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 = - 30.4 = - 120\).
b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).
\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)
\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)
a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)
\(=\dfrac{5}{7}\)
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
`@` `\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)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
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
? ( ͠° ͟ʖ ͡°)