thực hiện phép tính\(\left\{\left[3^{14}.69+3^{14}.12\right]:3^{16}-7\right\}:2^4\)
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HT
2
20 tháng 10 2017
\(\left\{\left[3^{14}.69+3^{14}.12\right]:3^{16}-7\right\}:2^4\\ =\left[3^{14}\left(69+12\right):3^{16}-7\right]:2^4=\left[\left(3^{14}.3^4\right):3^{16}-7\right]:2^4\\ =\left(3^2-7\right):2^4=\left(9-7\right):2^4=2:2^4=\dfrac{1}{8}\)
20 tháng 10 2017
Bạn ơi! Bạn viết sai dấu ngoặc nhé! Mình sửa thành:
[(314. 69 + 314. 12) : 316 - 7] : 24
= {[ 314. (69 + 12)] : 316 - 7} : 24
= {[ 314. 81] : 316 - 7} : 24
= {[314. 34] : 316 - 7} : 24
= {318 : 316 - 7} : 24
= {32 - 7} : 24
= {9 - 7} : 24
= 2 : 16
= \(\dfrac{1}{8}\)
= 8.
PN
20 tháng 5 2017
a) \(\left(-28\right).\left(-3\right).\left(+4\right).\left(-7\right)=\left(+84\right).\left(+4\right).\left(-7\right)=\left(+336\right).\left(-7\right)=\left(-2352\right)\)
b) \(2.8.\left(-14\right).\left(-3\right)=16.\left(-14\right).\left(-3\right)=\left(-224\right).\left(-3\right)=672\)
VB
1
S
29 tháng 12 2020
\(\dfrac{20^3.\left(-49\right)^2}{14^3.5^4}=\dfrac{2^3.2^3.5^3.7^4}{2^3.7^3.5^4}=\dfrac{2^3.7}{5}=\dfrac{8.7}{5}=\dfrac{56}{5}\)
1 tháng 2 2019
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
15 tháng 3 2018
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
AH
Akai Haruma
Giáo viên
31 tháng 12 2020
Lời giải:
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)
\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)
3 tháng 7 2023
Sửa đề: (2/7)^7*7^7
\(A=\dfrac{\left(2\right)^7+\left(\dfrac{9}{3}:\dfrac{3}{16}\right)^3}{2^7\left(5^2+2^2\right)}\)
\(=\dfrac{\left(2\right)^7+\left(16\right)^3}{2^7\cdot29}\)
\(=\dfrac{2^7+2^7\cdot2^5}{2^7\cdot29}=\dfrac{1+2^5}{29}=\dfrac{33}{29}\)
\(\left\{\left[3^{14}.69+12.3^{14}\right]:3^{16}-7\right\}:2^4\)
\(=\left\{\left[3^{14}.\left(69+12\right)\right]:3^{16}-7\right\}:16\)
\(=\left\{\left[3^{14}.81\right]:3^{16}-7\right\}:16\)
\(=\left\{\left[3^{14}.3^4\right]:3^{16}-7\right\}:16\)
\(=\left\{3^{18}:3^{16}-7\right\}:16\)
\(=\left\{3^2-7\right\}:16\)
\(=\left\{9-7\right\}:16\)
\(=2:16\)
\(=\frac{1}{8}\)