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ai làm cho mình bài này với:
tính:A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
mình tích cho 3 luôn
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7.11}{4}.\left(\frac{1}{4}+\frac{3}{20}+\frac{1}{10}+\frac{1}{14}\right)\)
\(A=\frac{77}{4}.\left(\frac{35}{140}+\frac{21}{140}+\frac{14}{140}+\frac{10}{140}\right)\)
\(A=\frac{77}{4}.\frac{80}{140}\)\(=\frac{77}{8}.\frac{20}{35}\)
\(A=11\)
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}\cdot\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\left(\frac{1}{3}-\frac{1}{7}\right)\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\frac{4}{21}\right)\)
\(A=\frac{7}{4}\cdot\frac{132}{21}=11\)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
A= -1 - (2-3-4+5) - (6+7+8-9) -... - (198 -199-120+121)
A= -1 - 0-0-...-0
A= -1
nhớ k giùm mk
a)= 21 + (-21)
= 0
b)=15+4.3
=15+12
=27
c) =8.3+36:12-27
=24+3-27
=27-27
=0
tui ko biết
và đừng hỏi tui vì sao
vì tui học lớp 5 =_=
ok nhá
a)A=1+2+22 +23+24 +...+259
2A=2+22+23+24+...+259+260
2A - A=(2+22+23+24+...+259+260)-(1+2+22 +23+24 +...+259)
A=260-1
b)A=1+2+22 +23+24 +...+259(60 số hạng)
=(1+2+22)+(23+24 +25)+...+(258+259+260)
=1x(1+2+22)+23x(1+2+22)+...+258x(1+2+22)
=1x7+23x7+...+258x7
=(1+23+...+258)x7 chia hết cho
Vậy A chia hết cho 7(Đpcm)
Học tốt
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
\(a,x-\left(-15\right)=-8\\ =>x+15=-8\\ =>x=-8-15\\ =>x=-23\\ b,-40-x=-35\\ =>x=\left(-40\right)-\left(-35\right)\\ =>x=-40+35\\ =>x=-5\\ c,x+\left(-50\right)=-27\\ =>x-50=-27\\ =>x=-27+50\\ =>x=23\)
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:
\(a,x-\left(-15\right)=-8\)
\(x=-8+\left(-15\right)\)
\(x=-23\)
\(b,-40-x=-35\)
\(x=-40-\left(-35\right)\)
\(x=-5\)
\(c,x+\left(-50\right)=-27\)
\(x=-27-\left(-50\right)\)
\(x=23\)
A=64.8.1/1024
A=512.1/1024
A=1/2
A=0.5