a: \(=\dfrac{2\cdot136-28\cdot6+62\cdot3}{30}\cdot\dfrac{7}{8}=\dfrac{290}{30}\cdot\dfrac{7}{8}=\dfrac{29}{3}\cdot\dfrac{7}{8}=\dfrac{203}{24}\)

b: \(=\dfrac{3}{11}\cdot\dfrac{4}{11}+\dfrac{3}{13}\cdot\dfrac{4}{11}-\dfrac{1}{13}\)

\(=\dfrac{4}{11}\left(\dfrac{3}{11}+\dfrac{3}{13}\right)-\dfrac{1}{13}\)

\(=\dfrac{4}{11}\cdot\dfrac{72}{143}-\dfrac{1}{13}=\dfrac{167}{1573}\)

\(a)A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}\)

\(=\frac{(23+1)\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}+\frac{3}{11}+\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}+\frac{9}{13}+\frac{9}{7}+\frac{9}{11}+9}=\frac{47-23+24}{47-23+24}\cdot\frac{3(1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13})}{3(3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11})}\)

\(=\frac{1+\frac{1}{7}+\frac{1}{11}+\frac{1}{1001}+\frac{1}{13}}{3+\frac{3}{1001}+\frac{3}{13}+\frac{3}{7}+\frac{3}{11}}=\frac{1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11}}{3(1+\frac{1}{1001}+\frac{1}{13}+\frac{1}{7}+\frac{1}{11})}=\frac{1}{3}\)

\(b)\)\(\text{Đặt A = }1+2+2^2+2^3+...+2^{2012}\)

\(2A=2(1+2^2+2^3+...+2^{2012})\)

\(2A=2+2^2+2^3+...+2^{2013}\)

\(2A-A=(2+2^2+2^3+2^4+...+2^{2013})-(1+2+2^2+2^3+...+2^{2012})\)

\(\Rightarrow A=2^{2013}-1\)

\(\text{Quay lại bài toán,ta có :}\)

\(B=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2(2^{2013}-1)}=\frac{1}{2}\)

bài 1b)

\(8\frac{1}{14}-6\frac37\)

C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)

C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)

\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)

bài 1 c)\(7-3\frac67\)

C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)

C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)

BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

\(\frac{11}{12}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{11}{12}.\frac{1}{3}=\frac{11}{36}\)

Mình​ cũng đang k pit lm câu này

​

A=(24.47-23)/(24+47-23) . [3(1+1/7-1/11-1/13+1/1001)]/[9(1+1/7-1/11-1/13+1/1001)]

=1105/48 . 3/9 =1105/144

- Vì bạn giải xong nên mình đưa kết quả bạn soát nha :

\(a,x=-\frac{1}{4}.b,x=-\frac{19}{5}.c,x=\frac{11}{5}.d,x=-\frac{134}{81}.e,x=\frac{271}{75}\)

\(g,x=\frac{3}{22}.h,x=\frac{6}{13}.i,x=\frac{1}{3}.k,\frac{5684}{837}\)

- Thiếu đề bài bạn ơi !!

- Nếu là giải phương trình thì nghĩ bạn nên tự làm bởi toàn là chuyển vế cơ bản ỷ lại quá sẽ khó để làm đến phần thông hiểu, nâng cao ,.... nha

- Nếu cần kết quả để xem đúng hay sai thì ấn hỏi mình hoặc các bạn khác cũng được nha !

a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)

\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\) 

\(A=\frac{1105}{28}.\)\(\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{9+\frac{9}{7}-\frac{9}{11}+\frac{9}{1001}-\frac{9}{13}}\)

\(A=\frac{1105}{28}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}\)

\(A=\frac{1105}{28}.\frac{3}{9}\)

\(A=\frac{1105}{84}\)

b)\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)

Đặt \(A=1+2+2^2+2^3+...+2^{2012}\)

Suy ra \(2.A=2+2^2+2^3+2^4+...+2^{2013}\)

Khi đó \(2.A-A=2^{2013}-1\)hay \(A=2^{2013}-1\)

Do đó : \(M=\frac{A}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2014}-2}=\frac{1}{2}\)

          Vậy \(M=\frac{1}{2}\)

giúp em với : Akai Haruma