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#)Giải ;
b) Đặt \(N=1+2+2^2+2^3+...+2^{2012}\)
\(\Rightarrow2N=2+2^2+2^3+2^4+...+2^{2013}\)
\(\Rightarrow2N-N=N=\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(\Rightarrow N=2^{2013}-1\)
Thay N vào M, ta có :
\(M=\frac{2^{2013}-1}{2^{2014}-2}\)
Thêm Cho pen
\(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
Phải tính hết nhé
\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}+\frac{3}{11}-\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}+9}\)
\(A=\frac{1105}{48}.\frac{3.\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{1001}+\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{11}+\frac{1}{7}+1\right)}\)
\(A=\frac{1105}{48}.\frac{3.\frac{1311}{1001}}{9.\frac{1054}{1001}}\)
\(A=\frac{1105}{48}.\frac{3933}{1001}:\frac{9468}{1001}\)
\(A=\frac{1105}{48}.\frac{437}{1052}\)
\(A\approx9,56\)
Chú ý : Dấu xấp xỉ \(\approx\)
\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)
\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)
\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)
\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)
\(2A=2(1+2^2+2^3+...+2^{2012})\)
\(2A=2+2^2+2^3+...+2^{2013}\)
\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)
\(\Rightarrow A=2^{2013}-1\)
\(\text{Quay lại bài toán,ta có :}\)
\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)
bó tay luôn