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\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
a . Ta có : \(A=\frac{25^3.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
b . Ta có : \(B=\frac{2^6.6^3}{8^2.9^2}=\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
c . Ta có :
\(C=\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}=\frac{\left(3.5\right)^3+5.\left(3.5\right)^2-5^3}{\left(6.3\right)^3+6.\left(3.6\right)^2-6^3}=\frac{5^3.\left(3^3-3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}=\frac{5^3}{6^3}\)
d . Ta có :
\(D=\frac{\left(7^4-7^3\right)^2}{49^3}=\frac{7^4-7^3}{49^3}.\left(7^4-7^3\right)=\left(\frac{7^4}{49^3}-\frac{7^3}{49^3}\right).\left(7^4-7^3\right)\)
\(=\left(\frac{7^4}{7^6}-\frac{7^3}{7^6}\right).\left(7^4-7^3\right)=\left(\frac{1}{7^2}-\frac{1}{7^3}\right).\left(7^4-7^3\right)\)
\(=\frac{6}{7^3}.\left(7^4-7^3\right)=\frac{6}{7^3}.7^3.\left(7-1\right)=36\)
a) \(A=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
b) \(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^5.2^3.3^3}{2^6.3^4}=\frac{2^8.3^3}{2^6.3^4}=\frac{2^2}{3}=\frac{4}{3}\)
1)
b) \(\left(-0,25\right):\left(\frac{1}{2}\right)^2\)
\(=\left(-\frac{1}{4}\right):\frac{1}{4}\)
\(=-1.\)
h) \(\left(-15\right).0,23+\left(-15\right).0,77\)
\(=\left(-15\right).\left(0,23+0,77\right)\)
\(=\left(-15\right).1\)
\(=-15.\)
Chúc bạn học tốt!
15^3+5.15^2-5^3 / 18^3+6.18^2-6^3
=3375+5.225-125 /5832+6.324-216
=3375+1125-125 /5832+1944-216
=4500-125 /7826-216
=4375 /7610
=875 /1522
mk ko bt viết phần nên viết phần bằng dấu gạch / nhé.mk cx ko bt đúng hay sai đâu nhé
Gọi giá trị trên là : A
\(A=3^{100}-3^{99}+3^{98}+....+3^2-3+1\)
\(\Rightarrow3A=3^{101}-3^{100}+3^{99}-3^{98}+......+3^3-3^2+3\)
\(\Rightarrow3A+A=3^{101}+1\)
\(\Rightarrow4A=3^{101}+1\Rightarrow A=\frac{3^{101}+1}{4}\)
a) \(\left|\frac{1}{3}x-8\right|+3=15\)
\(\Leftrightarrow\left|\frac{1}{3}x-8\right|=12\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x-8=-12\\\frac{1}{3}x-8=12\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x=-4\\\frac{1}{3}x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=60\end{cases}}\)
Vậy \(x\in\left\{-12;60\right\}\)
b) \(15-\left|2+3x\right|=8\)
\(\Leftrightarrow\left|2+3x\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=-7\\2+3x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-9\\3x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{5}{3}\end{cases}}\)
Vậy \(x\in\left\{-3;\frac{5}{3}\right\}\)
d) \(-1\frac{1}{6}-\left|5-3x\right|=\frac{2}{3}\)
\(\Leftrightarrow\frac{-7}{6}-\left|5-3x\right|=\frac{2}{3}\)
\(\Leftrightarrow\left|5-3x\right|=\frac{-7}{6}-\frac{2}{3}\)
\(\Leftrightarrow\left|5-3x\right|=\frac{-11}{6}\)
Vì \(\left|5-3x\right|\ge0\forall x\)
mà \(\frac{-11}{6}< 0\)\(\Rightarrow\)Vô lý
Vậy \(x\in\varnothing\)
e) \(\left(\frac{3}{7}\right)^{20}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{20}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{2.6}\)
\(=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^8\)
g) \(4.2^5:\left(2^3.1^{16}\right)=2^2.2^5:2^3=2^4=16\)
sorry mk chỉ giải đk câu a thui nhé
(2.5)^2= =100 ; (6.5)^3 =27000
k cho mk vs dù sao thì mk cũng giúp đk bn mà ^ ^