\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}+\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)\)\(+....+\frac{1}{x}\left(1+2+3+...+x\right)\)

   \(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x\left(x+1\right)}{2}\)

   \(=\frac{1}{2}\left(2+3+4+...+\left(x+1\right)\right)\)

   \(=\frac{1}{2}.\frac{\left[\left(x+1\right)+2\right]x}{2}\)

   \(=\frac{1}{4}\left(x+3\right)x\)

\(B=115\)

\(\Leftrightarrow\frac{1}{4}.x\left(x+3\right)=115\)

\(\Leftrightarrow x\left(x+3\right)=115.4\)

\(\Leftrightarrow x\left(x+3\right)=20.23\)

\(\Leftrightarrow x=20\)

Vậy....

Bạn ơi dạy mình cách tính dong thứ 3 dấu = thứ nhất đấy phân tích kiểu nào cho nhanh vậy

Bạn ơi với lại chỗ \(x\left(x+3\right)=20.23\)

Tính kiểu nào ra 20 vậy

Mình vừa giải xong cho bạn khác ở đây: https://hoc24.vn/hoi-dap/question/866070.html . Thấy câu hỏi của bạn trùng với bạn trước, đây là phần bài giải của mình nhé!

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+3+...+x\right)\)

\(B=1+\frac{1}{2}\left(1+2\right)\cdot2:2+\frac{1}{3}\left(1+3\right)\cdot3:2+...+\frac{1}{x}\left(1+x\right)\cdot x:2\)

\(B=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+x}{2}\)

\(B=1+\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}\)

De B = 115

=> \(\frac{\left(1+1+...+1\right)+\left(2+3+...+x\right)}{2}=114\)

=> (1 + 1 + ... + 1) + (2 + 3 + ... + x) = 228

den day chju :v

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.............+\frac{1}{x}\left(1+2+3+............+x\right)\)

\(=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}\frac{3.4}{2}+...........+\frac{1}{x}\frac{x\left(x+1\right)}{2}\)

\(=\frac{1}{2}\left(2+3+4+.............+\left(x+1\right)\right)\)

\(=\frac{1}{2}\frac{\left[\left(x+1\right)+2\right]x}{2}\)

\(=\frac{1}{4}\left(x+3\right)x\)

\(B=115\Leftrightarrow\frac{1}{4}.x\left(x+3\right)=115\)

\(\Leftrightarrow x\left(x+3\right)=115.4\)

\(\Leftrightarrow x\left(x+3\right)=20.23\)

\(\Leftrightarrow x=20\)

Let me guess......?!

Ta có :

\(B=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)\)

\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)

\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)

\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)

để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)

\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)

\(\Rightarrow\)\(\left(x+3\right)x=23.20\)

\(\Rightarrow\)x = 20

3. a) \(đk:x\ne1;x\ne-2\)

Ta có: \(A=\frac{3x-3+2}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)

Để A là số nguyên thì x là số nguyên và x-1 là ước của 2 . Ta có bảng:

Lại có: \(B=\frac{2x^2+4x-3x-6+5}{x+2}=\frac{2x\left(x+2\right)-3\left(x+2\right)+5}{x+2}=2x-3+\frac{5}{x+2}\)

Để B là số nguyên thì x là số nguyên và x+2 là ước của 5. Ta có bảng:

b) Để A và B cùng nguyên thì \(x\in\left\{-1;3\right\}\)

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Các bạn ơi giải giúp mink vs mink đg cần gấp

nhiều thế ??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)

\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)

\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)