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\(\dfrac{1}{a+1}+\dfrac{1}{a\left(a+1\right)}=\dfrac{a\left(a+1\right)}{\left(a+1\right)a\left(a+1\right)}+\dfrac{a+1}{a\left(a+1\right)\left(a+1\right)}\\ =\dfrac{a+1+a\left(a+1\right)}{a\left(a+1\right)\left(a+1\right)}=\dfrac{\left(a+1\right)\left(a+1\right)}{a\left(a+1\right)\left(a+1\right)}\\ =\dfrac{1}{a}\)
Suy ra \(\dfrac{1}{a}=\dfrac{1}{a+1}+\dfrac{1}{a\left(a+1\right)}\)
Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a\left(a+1\right)}\)
\(=\dfrac{1}{a\left(a+1\right)}\) (đpcm)
Ta được:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...-\dfrac{1}{100}\) \(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)
b,
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)
\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)
\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)
\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
Bài 1:
Ta có:
\(\left(100a+3b+1\right)\left(2^a+10a+b\right)=225\left(1\right)\)
Mà \(225\) lẻ nên \(\left\{{}\begin{matrix}100a+3b+1\\2^a+10a+b\end{matrix}\right.\) cùng lẻ \(\left(2\right)\)
\(*)\) Với \(a=0\) ta có:
Từ \(\left(1\right)\Leftrightarrow\left(100.0+3b+1\right)\left(2^a+10.0+b\right)=225\)
\(\Leftrightarrow\left(3b+1\right)\left(1+b\right)=225=3^2.5^2\)
Do \(3b+1\div3\) dư \(1\) và \(3b+1>1+b\)
Nên \(\left(3b+1\right)\left(1+b\right)=25.9\) \(\Rightarrow\left\{{}\begin{matrix}3b+1=25\\1+b=9\end{matrix}\right.\) \(\Leftrightarrow b=8\)
\(*)\) Với \(a\ne0\left(a\in N\right)\) ta có:
Khi đó \(100a\) chẵn, từ \(\left(2\right)\Rightarrow3b+1\) lẻ \(\Rightarrow b\) chẵn
\(\Rightarrow2^a+10a+b\) chẵn, trái với \(\left(2\right)\) nên \(b\in\varnothing\)
Vậy \(\left\{{}\begin{matrix}a=0\\b=8\end{matrix}\right.\)
Bài 2:
Ta có:
\(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+...+2017}\)
\(=\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}+\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}+...+\dfrac{1}{\dfrac{\left(1+2017\right).1009}{2}}\)
\(=\dfrac{2}{2.4}+\dfrac{2}{3.6}+\dfrac{2}{4.8}+...+\dfrac{2}{1009.2018}\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{1009.1009}\)
\(\Rightarrow A< \dfrac{1}{2.2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1008.1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\) (Đpcm)
2
a. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{2}-\dfrac{1}{100}\)
=\(\dfrac{49}{100}\)
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
1a.Vì \(\left|x\right|\) là 1 số tự nhiên nên \(\left|x\right|+2017\ge2017\)(1)
Mà ta đã biết:\(\dfrac{a}{b}\ge\dfrac{a}{b+n}\)với n là một số tự nhiên.
Nên từ (1)suy ra\(\dfrac{2016}{\left|x\right|+2017}\le\dfrac{2016}{2017}\)
Vậy để \(\dfrac{2016}{\left|x\right|+2017}\)lớn nhất thì \(\dfrac{2016}{\left|x\right|+2017}=\dfrac{2016}{2017}\)
1b.Ta thấy:
\(\dfrac{\left|x\right|+2016}{-2017}=\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)
Để \(\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)lớn nhất thì \(-\left(\left|x\right|+2016\right)\)lớn nhất
Mà theo câu a,ta có:\(\left|x\right|\)+2016 là một số tự nhiên nên \(-\left(\left|x\right|+2016\right)\)mang dấu âm hay \(-\left(\left|x\right|+2016\right)\le0\)( chú ý \(-0=0\))
Vậy để \(-\left(\left|x\right|+2016\right)\)lớn nhất hay \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì \(\left|x\right|+2016=0\)
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì nó bằng \(\dfrac{0}{-2017}\)hay nó bằng 0
2)
a)Để \(\dfrac{\left|x\right|+1945}{1975}\)nhỏ nhất thì \(\left|x\right|+1945\) nhỏ nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1945\ge1945\)
\(\Rightarrow\)Để \(\left|x\right|+1945\) nhỏ nhất thì \(\left|x\right|+1945\) = 1945
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+1945}{1975}\)bé nhất thì nó phải bằng \(\dfrac{1945}{1975}\)hay\(\dfrac{389}{395}\)
b)Để \(\dfrac{-1}{\left|x\right|+1}\)thì \(\left|x\right|+1\)bé nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1\ge1\)
\(\Rightarrow\)Để \(\left|x\right|+1\)bé nhất thì \(\left|x\right|+1\)\(=1\)
\(\Rightarrow\)GTNN của \(\dfrac{-1}{\left|x\right|+1}\)là \(\dfrac{-1}{1}\) hay -1
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
quy đồng 1/a + (-1/(a+1) = 1/a(a+1) + (-a/a(a+1) = (a+1-a)/a(a+1) = 1/a(a+1)
Ta có:
VP = 1/a + [-1/(a + 1)]
= (a + 1 - a)/[a.(a + 1)]
= 1/[a(a + 1)]
= VT
Vậy 1/[a.(a + 1)] = 1/a + [-1/(a + 1)]