a.4^7

b.8^5 

 c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

1: \(=\dfrac{272-168+186}{30}\cdot\dfrac{7}{9}=\dfrac{29}{3}\cdot\dfrac{7}{9}=\dfrac{203}{27}\)

2: \(=\dfrac{9-55}{33}\cdot\dfrac{6-5}{8}-\dfrac{4}{5}\cdot\dfrac{1}{24}=\dfrac{-4}{3}\cdot\dfrac{1}{8}-\dfrac{1}{30}=\dfrac{-1}{6}-\dfrac{1}{30}=\dfrac{-6}{30}=-\dfrac{1}{5}\)

3: \(=\dfrac{7}{3}\left(\dfrac{6}{25}+\dfrac{19}{25}\right)=\dfrac{7}{3}\cdot1=\dfrac{7}{3}\)

bai nao vay

\(A=1.\left(-1\right)+3.\left(-1\right)^2+5.\left(-1\right)^3.7+\left(-1\right)^4+9.\left(-1\right)^5\)

\(A=1.\left(-1\right)+3.1+5.\left(-1\right).7+1+9.\left(-1\right)\)

\(A=\left(-1\right)+3+\left(-5\right).7+1+\left(-9\right)\)

\(A=-1+3-35+1-9\)

\(A=-41\)

Ý bạn là mỗi phép tính phải thêm dấu và ....để kết quả bằng 6 phải không ạ

ừ

ví dụ:(1+1+1)!=6

sửa đề: \(B=5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}\)

Ta có: \(A=\frac23+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}+\frac{194}{195}\)

\(=1-\frac13+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}+1-\frac{1}{143}+1-\frac{1}{195}\)

\(=7-\left(\frac13+\frac{1}{15}+\cdots+\frac{1}{195}\right)\)

\(=7-\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{13\cdot15}\right)\)

\(=7-\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{13}-\frac{1}{15}\right)=7-\frac12\left(1-\frac{1}{15}\right)\)

\(=7-\frac12\cdot\frac{14}{15}=7-\frac{7}{15}=\frac{98}{15}\) >6

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{7^2}<\frac{1}{6\cdot7}=\frac16-\frac17\)

Do đó; \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<1-\frac12+\frac12-\frac13+\cdots+\frac16-\frac17=1-\frac17<1\)

=>\(5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<5+1=6\)

=>B<6

mà A>6

nên B<A

b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)

                                                                                                       \(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

   Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)                                                    

Vậy x= 4