Giúp tôi giải toán và làm văn

Đặt \(\frac{x+y}{7}=\frac{x-y}{1}=\frac{xy}{24}=k\)

\(\Rightarrow x+y=7k;x-y=k;xy=24k\)

\(\Rightarrow x=\left(7k+k\right):2=4k\)

\(y=3k\)

\(\Rightarrow xy=12k^2=24k\)

\(\Rightarrow k^2=2k\)

\(\Rightarrow k=2\)

\(\Rightarrow x=4.2=8;y=2.3=6\)

Với \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

\(\Rightarrow D=2\left(2x-1\right)-3\left(2x+3\right)\)

\(\Rightarrow D=4x-2-6x-9\)

\(\Rightarrow D=-2x-11\)

Với \(2x+3< 0\Rightarrow x< -\frac{3}{2}\)

\(\Rightarrow D=2\left(1-2x\right)-3\left(2x+3\right)\)

\(\Rightarrow D=2-4x-6x-9\)

\(\Rightarrow D=-10x-7\)

Thanks nha!

\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^5}\)

\(\Rightarrow7A=1+\frac{1}{7}+...+\frac{1}{7^4}\)

\(\Rightarrow7A-A=1-\frac{1}{7^5}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^5}}{6}\)

\(7^50\) là cái gì????????

  A = 1 x 2 + 2 x 3 + 3 x 4 + ... + n x (n + 1)

3A = 3 ( 1 x 2 + 2 x 3 + 3 x 4 + ... + n x (n + 1) )

3A = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + n x (n + 1) x 3

3A = 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + n x (n + 1) x ((n + 2) - (n - 1))

3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + n x (n + 1) x (n + 2) - (n - 1) x n x (n + 1)

3A = n x (n + 1) x (n + 2)

  A = n x (n + 1) x (n + 2) : 3

\(A=1.2+2.3+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Sorry, đề bài thiếu: a,b,c,d là số dương