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\(A=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{8}{11}-\dfrac{5}{7}.\dfrac{2}{11}\)
\(A=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{8}{11}-\dfrac{2}{11}\right)\)
\(A=\dfrac{5}{7}.\dfrac{5+8-2}{11}\)
\(A=\dfrac{5}{7}.\dfrac{11}{11}\)
\(A=\dfrac{5}{7}.1=\dfrac{5}{7}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{63}\)
\(B=\dfrac{95}{72}\)
\(C=\dfrac{4^6.9^5+6^9.120}{8^4-3^{12}-6^{11}}\)
\(C=\dfrac{\left(2^2\right)^3.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(C=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(C=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(C=\dfrac{2.6}{5.3}=\dfrac{12}{15}=\dfrac{4}{5}\)
Câu 1 xem kỉ đề
\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)
a) A=212.35-\(\frac{2^{12}.3^6}{2^{12}}\)+93+84.35
=212.35-36+36+212.35
=213.35
b)B=496.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)
=496.5-7.5-2.495
=712.5-7.5-2.710
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
