Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
7.12^8.5^4-6^9.20^4=7.96.20-54.80=9120
6^10.10^5-7,3^9.80^4=60.50-65.7.320=3000
ket qua rut gon thanh dap so 456 phan 150
a)93/7:-3/7+178/7:3/7
3/7.(-93/7+178/7)=3/7.85/7=255/49
b)7.128.54-69.204/610.105-7.39.804
7.38.216.5429.39.54.28:210.310.25.55-7.39.216.54
=72.336.244.517
\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)
= 
= 
= 
= 
. 
a) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{17}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{13^2}{12^2}=\dfrac{169}{144}\)
b)\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2=\left(\dfrac{-1}{12}\right)^2=\dfrac{\left(-1\right)^2}{12^2}=\dfrac{1}{144}\)
c)\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^4.2^8}{5^{10}.2^{10}}=\dfrac{5^8.2^8}{5^8.5^2.2^8.2^2}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)
d)\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{\left(-10\right)^5.\left(-6\right)^4}{3^5.5^4}=\dfrac{\left(-2\right)^5.5^5.2^4.3^4}{3^4.3.5^4}=\dfrac{\left(-2\right)^5.5.5^42^4}{3.5^4}=\dfrac{\left(-2\right)^5.5.2^4}{3}=\dfrac{-2560}{3}=-853\dfrac{1}{3}\)
1,\(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3\) = \(2:\left(\dfrac{-1}{216}\right)\)=\(-432\)
2, \(a,2^6.5^6.5^4.4^4=10^6.20^4\)
\(b,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)
b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)
c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)
d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)
e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)
Bài 5: GTNN chứ nhỉ?
Với mọi gt của \(x;y\in R\) ta có:
\(x^2+3\left|y-2\right|+1\ge1\)
Hay \(A\ge1\) với mọi gt của \(x;y\in R\)
Dấu "=" sảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Vậy..................
Bài 6: GTLN chứ?
Với mọi giá trị của \(x\in R\) ta có:
\(-\left(2x-1\right)^2\le0\Rightarrow-5-\left(2x-1\right)^2\le-5\)
Hay \(B\le5\) với mọi giá trị của \(x\in R\)
Dấu "=" sảy ra khi và chỉ khi \(x=\dfrac{1}{2}\)
Vậy...................
Bài 4 :
\(a,3^{15}-9^6=3^{15}-\left(3^2\right)^6=3^{15}-3^{12}=3^{12}\left(3^3-1\right)=3^{12}.26=3^{12}.2.13⋮\left(đpcm\right)\)
\(b,8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}=2^{18}\left(2^3-1\right)=2^{18}.7=2^{17}.2.7=2^{17}.14⋮14\left(đpcm\right)\)
Bài 5 :
\(A=1^2+3^2+6^2+9^2+.............+39^2\)
\(=1+3^2+\left(6^2+9^2+.........+39^2\right)\)
\(=10+3^2\left(2^2+3^2+.........+13^2\right)\)
\(=10+3^2.818\)
\(=10+9.818\)
\(=7372\)
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
Ta có: \(\dfrac{7.12^8.5^4-6^9.20^4}{6^{10}.10^5-7.3^9.80^4}\)
\(=\dfrac{7.3^8.2^{16}.5^4-2^9.3^9.5^4.2^8}{2^{10}.3^{10}.2^5.5^5-7.3^9.2^{15}.2.5^4}\)
\(=\dfrac{2^{16}.5^4.3^8\left(7-2.3\right)}{2^{15}.3^9.5^4\left(15-7.2\right)}\)
\(=\dfrac{2^{15}.5^4.3^8.2}{2^{15}.3^8.3.5^4}\)
\(=\dfrac{2}{3}\).