Tính: \(\left(5^4+4^7\right)\left(8^9-2^7\right)\left(2^4-4^2\right)=?\) ? \(\left(7^{2003}+72002\right):7^{2001}\)=? \(\left(169-4^2\right).\left(169-5^2\right).....\left(169-13^2\right)\) =?
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\left(8+\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=8+\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=11+\dfrac{1}{2}+2\)
\(=\dfrac{27}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{21.273.1333.4161.10101}{91.651.2451.6643.14763}\)
\(=\frac{3.7.13.21.31.43.73.57.91.111}{7.13.21.31.43.57.73.91.111.133}=\frac{3}{133}\)
Tuy nhiên cách làm trên phải có máy tính mới làm đc:
Có thể sử dụng công thức:
\(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Sau đó phân h:
\(2^4+2^2+1=\left(2^2+2+1\right)\left(2^2-2+1\right)=7.3\)
\(4^4+4^2+1=\left(4^2+4+1\right)\left(4^2-4+1\right)=21.13\)
....Tiếp tực làm thì sẽ ra đc kết quả:
\(=\frac{3.7.13.21.31.43.73.57.91.111}{7.13.21.31.43.57.73.91.111.133}=\frac{3}{133}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) Ta thấy :
2^4 = 16
4^2 = 16
16 - 16 = 0
Số nào nhân với 0 cũng bằng 0 nên giá trị biểu thức trên là 0
b ) ( 7^2015 + 7^2014 ) : 7^2013
= 7^2015 : 7^2013 + 7^2014 : 7^2013
= 7^2 + 7
= 49 + 7
= 56
c ) ( 3 . 4 . 2^16 ) ^ 2 / 11 . 2^13 . 4^11 - 16^9
Tính phần mẫu trước .
11 . 2^13 . 4^11 - 16^9 = 11 . 2^13 . ( 2^2 ) ^11 - (2^4)^9 = 11 . 2^13 . 2^22 - 2^36 = 11. 2^35 - 2^36 = 11 . 2^35 - 2^35 . 2 = ( 11 - 2 ) . 2^35 = 9 . 2^35
Phần tử :
( 3 . 4 . 2^16 ) ^ 2 = 3^2 . ( 2^2 ) ^ 2 . ( 2^16 ) ^ 2 = 3 ^ 2 . 2^4 . 2^32 = 9 . 2^36
Vì các thừa số của mẫu và tử đều giống nhau nên có kết quả là 1 .
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lười làm qá, hì:
Hướng dẫn thôi nha.
B1: Phá bỏ ngoặc của các phép tính.
B2: Ghép những số nguyên vào vs nhau, phân số vào vs nhau
B3: Giao hoán những phân số có cùng mẫu để cộng vào, ở đây chỉ nói cộng vì trừ lp 7 là cộng vs số đối mà
B4: Tính hết ra là xong
B = (8+6-3) - (9/4-5/4-2/4) + (2/7-3/7-9/7)
B = 11 - 1/2 -10/7
B = 21/2 - 10/7
B = 127/14
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right)\)
\(=8-\frac{9}{4}+\frac{2}{7}+6+\frac{3}{7}-\frac{5}{4}-3-\frac{2}{4}+\frac{9}{7}\)
\(=11-4+2\)
=9