\(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\)=100x
Tìm x, biết \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+\left|x+\dfrac{1}{3\cdot4}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=100x\)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Tính: \(\left(1-\dfrac{2}{2\cdot3}\right)\left(1-\dfrac{2}{3\cdot4}\right)\left(1-\dfrac{2}{4\cdot5}\right)...\left(1-\dfrac{2}{99\cdot100}\right)\)
Tính nhanh:
A= \(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot100}\right)\)
Kiểm tra lại đề xem thừa số cuối có đúng quy luật của dãy không.
1.tính c = \(\left(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+...+18\cdot19\cdot20\cdot21\right)-5-10-15-...-200\)
2. tính d = \(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{18\cdot19\cdot20}-\dfrac{3}{1\cdot2}-\dfrac{3}{2\cdot3}-...-\dfrac{3}{19\cdot20}\)
3.tìm x: \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
4.tìm x: \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
giải nhanh giùm tớ nhé. ghi cả lời giả ra kìa
3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
Câu hỏi today là a, \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{101}{1540}\)
b,\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{2020}{2021}\)
Chúc các bn học tốt ❤
Giải:
a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540
1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540
1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540
1/3.(1/5-1/x+1)=101/1540
1/5-1/x+1=101/1540:1/3
1/5-1/x+1=303/1540
1/x+1=1/5-303/1540
1/x+1=1/308
⇒x+1=308
x=308-1
x=307
b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021
1/1-1/x+1=2020/2021
1/x+1=1/1-2020/2021
1/x+1=1/2021
⇒x+1=2021
x=2021-1
x=2020
Mk thấy đề bài hơi sai là:
1/x+(x+1) ➜ 1/x.(x+1)
mới ra đc kết quả!
1) Tìm x để phân thức sau bằng không:
\(\dfrac{x^3+x^2-x-1}{x^3+2x-3}\)
2) Tính tổng:
\(B=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
Làm theo hướng dẫn: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{k}+\dfrac{1}{k+2}\right)-\dfrac{1}{k+1}\)
Bài 1 :
Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy,.........
\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+.....+\left|x+\dfrac{1}{197.199}\right|=100x\)
Giải PT : \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
Lời giải:
Trong TH này ta thêm điều kiện $x$ là số nguyên dương.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)
Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)
\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)
\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)
\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)
Hiển nhiên ta thấy:
\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)
\(\sqrt{2017-x}\geq 0\)
\(2016-x\geq 0\)
Do đó pt trên vô nghiệm
Tức là không tìm đc $x$ thỏa mãn.
Tính tổng: \(B=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
Làm theo hướng dẫn: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{k}+\dfrac{1}{k+2}\right)-\dfrac{1}{k+1}\)
Lời giải:
Ta có: \(\frac{1}{k(k+1)(k+2)}=\frac{1}{2}.\frac{2}{k(k+1)(k+2)}=\frac{1}{2}.\frac{(k+2)-k}{k(k+1)(k+2)}\)
\(=\frac{1}{2}\left(\frac{k+2}{k(k+1)(k+2)}-\frac{k}{k(k+1)(k+2)}\right)=\frac{1}{2}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)
.......
\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}\)