a)\(\left(\frac{2}{3}+\frac{4}{5}+\frac{12}{7}\right)+\left(2^7.5^6-\frac{25^3}{\left(0.125\right)3}\right).\left(0.0001\right)^2\)
b) \(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
Những câu hỏi liên quan
2frac{1}{315}.frac{1}{651}-frac{1}{105}.3frac{650}{651}-frac{4}{315.651}+frac{4}{105}left(2+frac{1}{315}right).frac{1}{651}-frac{1}{105}.left(3+frac{650}{651}right)-frac{4}{315.651}+frac{4}{105}2.frac{1}{651}+frac{1}{315}.frac{1}{651}-frac{1}{105}.3+frac{1}{105}.frac{650}{651}-frac{4}{315.651}+frac{4}{105}frac{2}{651}+frac{1}{315.651}-frac{3}{105}-frac{650}{105.651}-frac{4}{315.651}+frac{4}{105}frac{2}{615}+left(frac{1}{315.651}-frac{4}{315.651}right)+left(frac{-3}{105}+frac{4}{105}right)-frac{6...
Đọc tiếp
\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)
cho mk hỏi làm sao từ \(A=2.\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
=> \(A=\left(2+\frac{1}{315}\right).\frac{1}{651}.......\)
Bạn chú ý rằng 315= 3.105 và 650=651-1 nha
Thế 2 cái đó vào A là ra thoi !!
Cho mình hỏi cái chỗ \(3\frac{650}{651}\)là hỗn số hay phép nhân thế
Roi mình giải lun cho
Xem thêm câu trả lời
Bài 1) Tính nanh nếu có thể1) 63^2-47^22) 127^2+146.127+73^23) 215^2-105^24) left(4+1right).left(4^2+1right).left(4^4+1right).left(4^6+1right)...left(4^{156}+1right)5) 2frac{1}{315}.frac{3}{651}-frac{3}{105}.3frac{650}{651}-frac{4}{315.651}+frac{4}{105}6) 444443.444448.444441-444445.4444440.444447
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Bài 1) Tính nanh nếu có thể
1) \(63^2-47^2\)
2) \(127^2+146.127+73^2\)
3) \(215^2-105^2\)
4) \(\left(4+1\right).\left(4^2+1\right).\left(4^4+1\right).\left(4^6+1\right)...\left(4^{156}+1\right)\)
5) \(2\frac{1}{315}.\frac{3}{651}-\frac{3}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
6) \(444443.444448.444441-444445.4444440.444447\)
1) \(63^2-47^2=\left(63+47\right)\left(63-47\right)=110.16=1760\)
2) \(127^2+146.127+73^2=\left(127+73\right)^2=200^2=40000\)
3) \(215^2-105^2=\left(215-105\right)\left(215+105\right)=110.320=35200\)
4) mk chỉnh lại đề:
\(\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^4-1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^{512}-1\right)\)
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Bài 1: Tính
S=\(\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{3}{315}.\left(4-\frac{1}{651}\right)-\frac{4}{315.651}+\frac{12}{315}\)
Bài 2:Tìm các số nguyên a và b sao cho \(\frac{2}{a-b}:\frac{a-3}{3\left(a-b\right)}\) được thương là số nguyên.
GIÚP MK VỚI
rút gona,frac{2.5^{22}-9.5^{21}}{25^{10}}b,frac{5left(3.7^{15}-19.7^{14}right)}{7^{16}+3.7^{15}}c,frac{left(left(-2right)^2right)^3.left(-4right)^2}{left(-2right)^3.left(-2right)^2} d,frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}e,2^3+3.left(frac{1}{8}right)^0-left(frac{1}{2^2}right).4+left[left(-2right)^2:frac{1}{2}right].8f,frac{left(frac{2}{5}right)^7.5^5+left(frac{9}{4}right)^3:left(frac{3}{16}right)^3}{2^7.5^2+512}
Đọc tiếp
rút gon
a,\(\frac{2.5^{22}-9.5^{21}}{25^{10}}\)
b,\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
c,\(\frac{\left(\left(-2\right)^2\right)^3.\left(-4\right)^2}{\left(-2\right)^3.\left(-2\right)^2}\)
d,\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
e,\(2^3+3.\left(\frac{1}{8}\right)^0-\left(\frac{1}{2^2}\right).4+\left[\left(-2\right)^2:\frac{1}{2}\right].8\)
f,\(\frac{\left(\frac{2}{5}\right)^7.5^5+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
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1.thực hiện phép tính
a.25.left(frac{-1}{5}right)^3+frac{1}{5}-2.left(frac{-1}{2}right)^2-frac{1}{2}
b.35^1_6:left(frac{-4}{5}right)-46^1_6:left(frac{-4}{5}right)
c.left(frac{-3}{4}+frac{2}{5}right):frac{3}{7}+left(frac{3}{5}+frac{-1}{4}right):frac{3}{7}
d.frac{7}{8}:left(frac{2}{9}-frac{1}{18}right)+frac{7}{8}.left(frac{1}{36}-frac{5}{12}right)
e.frac{1}{6}+frac{5}{6}.frac{3}{2}-frac{3}{2}+1
f.left(-0,75-frac{1}{4}right):left(-5right)+frac{1}{15}-left(frac{-1}{5}right):left(-3right)
Đọc tiếp
1.thực hiện phép tính
a.\(25.\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2.\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
b.\(35^1_6:\left(\frac{-4}{5}\right)-46^1_6:\left(\frac{-4}{5}\right)\)
c.\(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
d.\(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}.\left(\frac{1}{36}-\frac{5}{12}\right)\)
e.\(\frac{1}{6}+\frac{5}{6}.\frac{3}{2}-\frac{3}{2}+1\)
f.\(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(\frac{-1}{5}\right):\left(-3\right)\)
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
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\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
thực hiện các phép tính sau:
\(a,\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(b,\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)\)
\(c,-\frac{2}{3}\left(8\frac{1}{2}-7.75\right):\frac{1}{3}+0.125:2\frac{2}{3}\)
3frac{1}{2}-4frac{2}{3}+left[frac{3}{4}-2frac{1}{3}right]-left(frac{5}{6}-frac{7}{4}right)+5frac{1}{2}-32frac{2}{3}-1frac{2}{5}+1frac{3}{10}-left(frac{2}{5}-frac{5}{6}right)+frac{4}{15}-1frac{1}{3}left[2frac{1}{3}-1frac{4}{3}right]-left(frac{5}{4}-frac{7}{12}+frac{-11}{6}right)+frac{4}{3}-frac{3}{4}-3frac{3}{2}+5frac{4}{3}-left(frac{7}{6}-1frac{3}{4}right)+left[frac{2}{3}-2frac{1}{4}right]2frac{2}{3}-frac{5}{12}-left(1frac{3}{4}-2frac{1}{4}right)-left[1-1frac{1}{6}right]+left[frac{-5}{3}right]1f...
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\(3\frac{1}{2}-4\frac{2}{3}+\left[\frac{3}{4}-2\frac{1}{3}\right]-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)
\(2\frac{2}{3}-1\frac{2}{5}+1\frac{3}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-1\frac{1}{3}\)
\(\left[2\frac{1}{3}-1\frac{4}{3}\right]-\left(\frac{5}{4}-\frac{7}{12}+\frac{-11}{6}\right)+\frac{4}{3}-\frac{3}{4}\)
\(-3\frac{3}{2}+5\frac{4}{3}-\left(\frac{7}{6}-1\frac{3}{4}\right)+\left[\frac{2}{3}-2\frac{1}{4}\right]\)
\(2\frac{2}{3}-\frac{5}{12}-\left(1\frac{3}{4}-2\frac{1}{4}\right)-\left[1-1\frac{1}{6}\right]+\left[\frac{-5}{3}\right]\)
\(1\frac{1}{3}-5\frac{1}{2}-\left[\frac{5}{6}-2\frac{2}{3}\right]+\left[\frac{7}{12}-\frac{5}{6}\right]\)
\(\frac{8}{15}-\left(\frac{2}{5}-3\frac{1}{3}+\left[\frac{-5}{6}\right]\right)+\left[\frac{1}{2}-\frac{4}{5}\right]-\left(\frac{1}{6}-1\frac{1}{3}\right)\)
\(-2\frac{3}{2}+\left[\frac{5}{6}-1\frac{1}{3}\right]-\left(\frac{5}{12}-\frac{7}{6}\right)+\left[\frac{4}{3}-3\frac{1}{4}\right]\)
\(\frac{9}{10}-1\frac{2}{5}-\left(\frac{5}{6}-3\frac{1}{2}\right)-\left[2\frac{1}{4}-5\frac{2}{36}\right]-\left[1-2\frac{1}{15}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(1\frac{1}{5}-\left(\frac{-9}{10}-2\frac{1}{2}+\frac{3}{4}\right)+\left[\frac{1}{5}-2\frac{1}{2}\right]+\frac{7}{10}-\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(2\frac{1}{3}-\left(5\frac{1}{2}-2\frac{2}{3}\right)+\left[1\frac{1}{6}-2\frac{1}{2}\right]-\frac{5}{12}+\left(\frac{1}{4}-\frac{1}{8}\right)\)
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