Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Thực hiện phép tính ở VP ta có:
a) \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)
VP bằng VT nên đẳng thức trên là đúng
b) \(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\)
VP bằng VT nên đẳng thức trên là đúng
a, \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)
Ta có:
\(VP=\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{\left(a+1\right)-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}=VT\)
\(\rightarrow\) đpcm
b, \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta có:
\(VP=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}-\dfrac{1}{a+1}+\dfrac{1}{a+2}\)(áp dụng câu a)
\(=\dfrac{1}{a}-\dfrac{2}{a+1}+\dfrac{1}{a+2}\)
\(=\dfrac{\left(a+1\right)-2a}{a\left(a+1\right)}+\dfrac{1}{a+2}=\dfrac{\left[\left(a+1\right)-2a\right]\left(a+2\right)}{a.\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{\left(1-a\right)\left(a+2\right)+a\left(a+1\right)}{a\left(a+1\right)\left(a+2\right)}=\dfrac{a+2-a^2-2a+a^2+a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=VT\)
Chúc bạn học tốt!!!
bai 1
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)
\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)
ta có : \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow\dfrac{1}{2}A=A-\dfrac{1}{2}A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow A=2.\left(\dfrac{1}{2}A\right)=1-2\left(\dfrac{1}{2}\right)^{100}< 1\left(đpcm\right)\)
\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2A=2\cdot\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\right]\)
\(2A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{99}\)
\(A=1-\left(\dfrac{1}{2}\right)^{99}< 1\left(đpcm\right)\)
Lời giải:
\(A=\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+...+(\frac{1}{2})^{98}+(\frac{1}{2})^{99}\)
\(\Rightarrow 2A=1+\frac{1}{2}+(\frac{1}{2})^2+...+(\frac{1}{2})^{97}+(\frac{1}{2})^{98}\)
Trừ theo vế:
\(2A-A=1-(\frac{1}{2})^{99}\)
\(A=1-(\frac{1}{2})^{99}< 1\)
Ta có đpcm.
Lời giải:
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)
\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)
\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)
b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)
\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)
\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)
\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)
\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)
\(A=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2016}{2017}=\dfrac{1}{2017}\)
\(B=\dfrac{-3}{2}\cdot\dfrac{-4}{3}\cdot...\cdot\dfrac{-2018}{2017}=\dfrac{2018}{2}=1009\)
\(A\cdot B=\dfrac{1009}{2017}\)
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
a) \(\dfrac{a}{b}+\dfrac{-a}{b+1}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}+\dfrac{-a.b}{b\left(b+1\right)}=\dfrac{ab+a-ab}{b\left(b+1\right)}=\dfrac{a}{b\left(b+1\right)}\)
b) \(\dfrac{a}{b+1}+\dfrac{-a}{b}=\dfrac{ab}{b\left(b+1\right)}+\dfrac{-a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab-ab-a}{b\left(b+1\right)}=\dfrac{-a}{b\left(b+1\right)}\)
Giải:
a) Biến đổi VP, ta có:
\(\dfrac{1}{a}-\dfrac{1}{a+1}\)
\(=\dfrac{1.\left(a+1\right)}{a.\left(a+1\right)}-\dfrac{a.1}{a.\left(a+1\right)}\)
\(=\dfrac{a+1}{a.\left(a+1\right)}-\dfrac{a}{a.\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a.\left(a+1\right)}\)
\(=\dfrac{1}{a.\left(a+1\right)}\) (đpcm)
b) Biến đổi VP, ta được:
\(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{1\left(a+2\right)}{a\left(a+1\right)\left(a+2\right)}-\dfrac{1.a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{a+2-a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\) (đpcm)
Chúc bạn học tốt!!!