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\(a,\left(-7\right).\left(-7\right).\left(-7\right).\left(-7\right)\)
\(=\left(-7\right)^4=2401\)
\(b,\left(-3\right).\left(-3\right).\left(-3\right).\left(-4\right).\left(-4\right).\left(-4\right)\)
\(=\left(-3\right)^3.\left(-4\right)^3=12^3\)
\(c,4.9.25\)
\(=9.\left(4.25\right)=900=30^2\)
\(d,8.\left(-3\right)^3.\left(-125\right)\)
\(=\left(-3\right)^3.\left[\left( -125\right).8\right]\)
\(=\left(-3\right)^3.\left(-1000\right)\)
\(=\left(-3\right)^3.\left(-10\right)^3\)
\(=30^3\)
\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)
\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)
\(c.\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)
\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{3}\times\left[\left(\frac{-3}{4}+\frac{1}{4}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)
\(=\frac{7}{3}\times\frac{1}{2}\)
\(=\frac{7}{6}\)
\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)
\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)
\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)
\(=\frac{7}{8}\times\frac{44}{7}\)
\(=\frac{11}{2}\)
a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=\dfrac{35}{8}+\dfrac{17}{3}\)
\(=\dfrac{105}{24}+\dfrac{136}{24}\)
\(=\dfrac{241}{24}\)
b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)
\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)
\(=\dfrac{29}{8}+\dfrac{27}{7}\)
\(=\dfrac{419}{56}\)
c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)
\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)
\(=\dfrac{9}{8}+\dfrac{16}{3}\)
\(=\dfrac{155}{24}\)
d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)
\(=\dfrac{17}{6}:\dfrac{1}{2}\)
\(=\dfrac{17}{6}\cdot2\)
\(=\dfrac{17}{3}\)
e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{39}{4}-\dfrac{6}{7}\)
\(=\dfrac{249}{28}\)
a: =4+3/8+5+2/3
=9+9/24+16/24
=9+25/24
=216/24+25/24=241/24
b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)
=27/7+29/8
=419/56
c: =2+3/8-1-1/4+5+1/3
=6+3/8-1/4+1/3
=6+3/8+1/12
=144/24+9/24+2/24
=155/24
d: =(15/6+2/6):1/2
=17/6*2
=17/3
e: =(15/6-2/6)*9/2-6/7
=13/6*9/2-6/7
=117/12-6/7
=249/28
Câu 2;3;4 dễ quá... bỏ qua!!
Câu 5;6 khó quá ... khỏi làm!!
dễ quá bỏ qua!!, khó quá khỏi làm!!
cứ tiêu chí mày bạn sẽ vượt qua mọi bài toán... và nhanh chóng đạt 1đ.
a) \(\left(-5\right)\left(-5\right)\left(-5\right)\left(-5\right)\left(-5\right)=\left(-5\right)^5\)
b)\(\left(-2\right)\left(-2\right)\left(-2\right)\left(-3\right)\left(-3\right)\left(+3\right)=\left(-2\right)^3\cdot\left(-3\right)^2\cdot3=\left(-2\right)^3\cdot3^3\)
c) \(\left(-7\right)\cdot7\cdot5\cdot\left(-5\right)\left(-5\right)=\left(-7^2\right)\cdot\left(-5\right)^3\)
d) \(\left(-8\right)\left(-3\right)3\cdot125=\left(-2^3\right)\cdot\left(-3^2\right)\cdot5^3\)
e) \(27\cdot\left(-2\right)3\left(-7\right)\cdot49=3^4\cdot\left(-2\right)\cdot\left(-7^3\right)\)
a: \(\left(-5\right)\cdot\left(-5\right)\cdot\left(-5\right)\cdot\left(-5\right)\cdot\left(-5\right)=\left(-5\right)^5\)
b: \(\left(-2\right)\cdot\left(-2\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-3\right)\cdot\left(+3\right)=\left(-2\right)^3\cdot3^3\)
a, x + \(\dfrac{1}{5}\) = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{5}\)
x = \(\dfrac{19}{30}\)
b, x - \(\dfrac{3}{5}=\dfrac{6}{7}\)
x = \(\dfrac{6}{7}+\dfrac{3}{5}\)
x = \(1\dfrac{16}{35}\)
c, - x - \(\dfrac{7}{5}=\dfrac{-8}{9}\)
- x = \(\dfrac{-8}{9}+\dfrac{7}{5}\)
x = \(\dfrac{23}{45}\)
d, \(\dfrac{3}{8}-x=\dfrac{2}{3}\)
\(x=\dfrac{3}{8}-\dfrac{2}{3}\)
\(x=\dfrac{-7}{24}\)
e, \(|x-\dfrac{8}{9}|=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{8}{9}\)
\(x=\dfrac{-11}{9}\)
Hiệu giữa SBT mới và cũ là:
353 – 23 = 330
Hiệu số phần bằng nhau là:
3-1 = 2 phần
Số bị trừ cũ là: 330 : 2 = 165
Số trừ cũ là : 165- 23 = 142
Đáp án cần chọn là: B