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12 tháng 3 2020

\(\text{(-27).18+77.(-27)+(-27).5}\)

\(=\left(-27\right).\left(18+22+5\right)\)

\(=\left(-27\right).45\)

\(=-1215\)

\(\left(-2\right)^3+\left(29-37\right)-\left(-10\right)\)

\(=\left(-8\right)+\left(-8\right)+10\)

\(=-16+10\)

\(=-6\)

12 tháng 3 2020

cảm ơn các bạn và bạn linh nha

1 tháng 12 2023

1) $(-81)\cdot27-3^4\cdot(-77)$

$=(-81)\cdot27-81\cdot(-77)$

$=-81\cdot[27+(-77)]$

$=-81\cdot(27-77)$

$=-81\cdot(-50)$

$=81\cdot50$

$=4050$

2) $(-16)\cdot32-(-4)^2\cdot18$

$=(-16)\cdot32-16\cdot18$

$=-16\cdot(32+18)$

$=-16\cdot50$

$=-800$

3) $(-2)^4\cdot289-16\cdot189$

$=16\cdot289+16\cdot(-189)$

$=16\cdot[289+(-189)]$

$=16\cdot(289-189)$

$=16\cdot100$

$=1600$

4) $41\cdot(-18)+41\cdot(-81)-41$

$=1\cdot(-18)+41\cdot(-81)+41\cdot(-1)$

$=41\cdot[(-18)+(-81)+(-1)]$

$=41\cdot(-18-81-1)$

$=41\cdot(-99-1)$

$=41\cdot(-100)$

$=-4100$

5) $17\cdot(-37)-23\cdot37-46\cdot(-37)$

$=17\cdot(-37)+23\cdot(-37)-46\cdot(-37)$

$=-37\cdot(17+23-46)$

$=-37\cdot(40-46)$

$=-37\cdot(-6)$

$=37\cdot6$

$=222$

6) $-48+48\cdot(-78)+48\cdot(-21)$

$=48\cdot(-1)+48\cdot(-78)+48\cdot(-21)$

$=48\cdot[(-1)+(-78)+(-21)]$

$=48\cdot(-1-78-21)$

$=48\cdot(-79-21)$

$=48\cdot(-100)$

$=-4800$

7) $29\cdot(-13)+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot13+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot[13+27+(-14)]$

$=-29\cdot(40-14)$

$=-29\cdot26$

$=-754$

$\text{#}Toru$

26 tháng 12 2022

Giải

Bài 1

\(1,a,-40-37-\left(-29\right)\\ =-40-37+29\\ =-77+29=-48\\ b,27+\left(-36\right)-\left(-45\right)=27-36+45\\ =-9+45\\ =36\\ c,-125-\left[\left(-18\right)-125\right]\\ =-125+18+125\\ =\left(-125+125\right)+18\\ =18\\ d,140+\left[\left(-184\right)-140\right]\\ =140-184-140\\ =\left(140-140\right)-184\\ =0-184\\ =-184\)

Bài 2 

\(a,x-\left(-15\right)=-8\\ =>x+15=-8\\ =>x=-8-15\\ =>x=-23\\ b,-40-x=-35\\ =>x=\left(-40\right)-\left(-35\right)\\ =>x=-40+35\\ =>x=-5\\ c,x+\left(-50\right)=-27\\ =>x-50=-27\\ =>x=-27+50\\ =>x=23\)

26 tháng 12 2022

1, Thực hiện phép tính:

\(a,-40+\left(-37\right)-\left(-29\right)=-77-\left(-29\right)=-48\)

\(b,27+\left(-36\right)-\left(-45\right)=-9-\left(-45\right)=36\)

\(c,-125-\left[\left(-18\right)-125\right]=-125-\left(-143\right)=18\)

\(d,140+\left[\left(-184\right)-140\right]=140+\left(-324\right)=-184\)

2, Tìm x biết:

\(a,x-\left(-15\right)=-8\)

\(x=-8+\left(-15\right)\)

\(x=-23\)

\(b,-40-x=-35\)

\(x=-40-\left(-35\right)\)

\(x=-5\)

\(c,x+\left(-50\right)=-27\)

\(x=-27-\left(-50\right)\)

\(x=23\)

29 tháng 12 2019

29-(10+29)=x-(27-9)

29-10-29    =x-18

10               =x-18

10+18          =x

28                  =x

5 tháng 9 2019

a) \(9.x-2.x=\frac{6^{27}}{6^{25}}+\frac{48}{12}\)

\(\Leftrightarrow7x=6^2+4\)

\(\Leftrightarrow7x=36+4=40\)

\(\Leftrightarrow x=\frac{40}{7}\)

Vậy : \(x=\frac{40}{7}\)

b) \(11^x=5.x+\frac{5^{31}}{5^{29}}+3.2^2-10^0\)

\(\Leftrightarrow11^x=5x+5^2+12-1\)

\(\Leftrightarrow11^x=5x+36\)

\(\Rightarrow x\in\varnothing\)

22 tháng 8 2020

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

22 tháng 8 2020

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)