a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

Kotori Minami bạn có thể giải chi tiết ra duoc ko

co ai biet ko

\(5\cdot2^{n+1}\cdot2^{-2}-2^n=384\)

\(5\cdot2^n\cdot2^{-1}-2^n=384\)

\(2^n\left(\frac{5}{2}-1\right)=384\)

\(2^n\cdot\frac{3}{2}=384\)

\(2^n=256\)

\(n=8\)

Câu hỏi của nguyễn quốc tú - Toán lớp 7 - Học toán với OnlineMath

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Có: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

 \(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

Hà mày cx tìm cái này ak

e có chơi ngọc rồng ko

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Có: \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

\(frac{x+4}{2000}\)