Câu 1:

\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4

Vậy A = 4

Câu 2:

\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)

\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)

\(B=1000-3.\left\{400-273\right\}\)

\(B=1000-3.127\)

\(B=1000-381\)

\(B=619\)

Vậy B = 619

B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)

=\(11\frac34-6\frac56+4\frac12+1\frac23\)

=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)

=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=10+\(\frac{13}{12}\)

=\(\frac{120}{12}+\frac{13}{12}\)

=\(\frac{133}{12}\)

b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)

= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)

=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)

=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)

=\(\frac{108}{60}\)

=\(\frac95\)

a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)

b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)

a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)

b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)

toàn hỏi lung tung. lớp 6 mà còn ko biết làm mấy bài toán vớ vẩn kia

A=4/63

a: =11+3/4-6-5/6+4+1/2+1+2/3

=10+9/12-10/12+6/12+8/12

=10+13/12=133/12

b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)

=3-11/15

=34/15

c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)

\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)

d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)

\(=4\cdot5-\dfrac{1+8}{3}\)

=20-3

=17

b: \(5^8:5^6+4\left(3^2-1\right)\)

\(=5^2+4\left(9-1\right)\)

=25+4*8

=25+32

=57

c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)

\(=400-36+20:\left[27-5\right]\)

\(=364+\dfrac{20}{22}\)

\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)

A) 2².5 - (1¹⁰ + 8) : 3²

= 4.5 - (1 + 8) : 9

= 20 - 9 : 9

= 20 - 1

= 19

B) 5⁸ : 5⁶ + 4.(3² - 1)

= 5² + 4.(9 - 1)

= 25 + 4.8

= 25 + 32

= 57

C) 400 - {36 - 20 : [3³ - (8 - 3)]}

= 400 - [36 - 20 : (27 - 5)]

= 400 -(36 - 20 : 22)

= 400 - (36 - 10/11)

= 400 - 386/11

= 4014/11

A=3/4

B=29/35

mình nghĩ thế

Sai thôi nhé!

\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)

\(A=1+\frac{1}{2}\)

\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)

\(A=\frac{3}{2}\)

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)