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Ta có:
\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)
\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)
\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Lại có:
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)
\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)
\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)
\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)
Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)
a) \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)
\(\Rightarrow\dfrac{1^m}{3^m}=\dfrac{1}{81}\)
\(\Rightarrow\dfrac{1}{3^m}=\dfrac{1}{3^4}\)
\(\Rightarrow m=4\)
b) \(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{9}{25}\right)^5\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left[\left(\dfrac{3}{5}\right)^2\right]^5\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
c) \(\left(-0,25\right)^p=\dfrac{1}{256}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{256}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{4^4}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\left(\dfrac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
2) a) \(\left(x+\dfrac{4}{5}\right)^2=\dfrac{9}{25}\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{3}{5}\\x+\dfrac{4}{5}=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{-7}{5}\end{matrix}\right.\) vậy \(x=\dfrac{-1}{5};x=\dfrac{-7}{5}\)
b) \(\left|x-\dfrac{3}{7}\right|=-2\) vì giá trị đối không âm được nên phương trình này vô nghiệm
c) điều kiện : \(x\ge-7\) \(\sqrt{x+7}-2=4\Leftrightarrow\sqrt{x+7}=4+2=6\)
\(\Leftrightarrow x+7=6^2=36\Leftrightarrow x=36-7=29\) vậy \(x=29\)
d) \(x^2-\dfrac{7}{9}x=0\Leftrightarrow x\left(x-\dfrac{7}{9}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-\dfrac{7}{9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{7}{9}\end{matrix}\right.\) vậy \(x=0;x=\dfrac{7}{9}\)
1) tìm GTNN
a) \(B=\left|x-2017\right|+\left|x-20\right|\)
B \(\ge\left|x-2017-x+20\right|=\left|-1997\right|=1997\)
Dấu " = " xảy ra khi và chỉ khi 20 \(\le x\le2017\)
Vậy MinB = 1997 khi 20 \(\le x\le2017\)
b) \(C=\left|x-3\right|+\left|x-5\right|\)
\(C\ge\left|x-3-x+5\right|=\left|2\right|=2\)
Dấu " = " xảy ra khi 3 \(\le x\le5\)
Vậ MinC = 2 khi và chỉ khi 3 \(\le x\le5\)
c) \(C=\left|x^2+4\right|+3\)
Ta thấy \(x^2+4\ge0\) với mọi x
nên \(\left|x^2+4\right|+3=x^2+4+3=x^2+7\)\(\ge\) 7
Dấu " =" xảy ra khi x = 0
MinC = 7 khi và chỉ khi x = 0
1)
a) \(0,25^x\cdot12^x=243\)
\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(38^y:19^y=512\)
\(\Leftrightarrow2y\cdot y=512\)
\(\Leftrightarrow2y^2=512\)
\(\Leftrightarrow y^2=256\)
\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)
Vậy \(y_1=-16;y_2=16\)
2)
a) \(3^x+3^{x+2}=2430\)
\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)
\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)
\(\Leftrightarrow10\cdot3^x=2430\)
\(\Leftrightarrow3^x=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(2^{x+3}-2^x=224\)
\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)
\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
3)
a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)
b) \(\left(x+0,7\right)^3=-27\)
\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x+\dfrac{3}{10}=-3\)
\(\Leftrightarrow x=-3-\dfrac{3}{10}\)
\(\Leftrightarrow x=-\dfrac{37}{10}\)
Vậy \(x=-\dfrac{37}{10}\)
4)
a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)
b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=1+1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
1. a) \(0,25^x.12^x=243\)
\(\Rightarrow\left(0,25.12\right)^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(38^y:19^y=512\)
\(\Rightarrow\left(38:19\right)^y=512\)
\(\Rightarrow2^y=2^9\)
\(\Rightarrow y=9\)
Vậy \(y=9.\)
2) a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy x=5.
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy x=5.
Bài 3: dễ tự làm.
Bài 1:
a) \(\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)......\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\)
= \(\dfrac{-8}{9}.\dfrac{-9}{10}.......\dfrac{-2003}{2004}.\dfrac{-2004}{2005}\) = \(\dfrac{-8}{2005}\)
b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\) = \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\)
= \(-2+\dfrac{1}{-2+\dfrac{1}{-1}}\) = \(-2+\dfrac{1}{-3}\) = \(\dfrac{-7}{3}\)
\(\text{Câu 1 : }\) Tính
\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)
\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)
a)\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1980}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1980}{1975}\\z=2004\end{matrix}\right.\)
b) \(\left|\dfrac{3}{4}+x\right|+\left|-\dfrac{1}{5}+y\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{4}+x\right|=0\\\left|-\dfrac{1}{5}+y\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
Lời giải:
Ta có \(A=\frac{1}{1.1981}+\frac{1}{2.1982}+...+\frac{1}{25.2005}\)
\(\Rightarrow 1980A=\frac{1980}{1.1981}+\frac{1980}{2.1982}+...+\frac{1980}{25.2005}\)
\(\Leftrightarrow 1980A=\frac{1981-1}{1.1981}+\frac{1982-2}{2.1982}+....+\frac{2005-25}{25.2005}\)
\(\Leftrightarrow 1980A=1-\frac{1}{1981}+\frac{1}{2}-\frac{1}{1982}+...+\frac{1}{25}-\frac{1}{2005}\)
\(1980A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+..+\frac{1}{2005}\right)\) (1)
Lại có:
\(25B=\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{1980.2005}\)
\(\Leftrightarrow 25B=\frac{26-1}{1.26}+\frac{27-2}{2.27}+...+\frac{2005-1980}{1980.2005}\)
\(\Leftrightarrow 25B=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{1980}-\frac{1}{2005}\)
\(25B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1980}\right)-\left(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{2005}\right)\)
\(25B=\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{1981}+\frac{1}{1982}+...+\frac{1}{2005}\right)\) (2)
Từ \((1); (2)\Rightarrow 1980A=25B\Rightarrow \frac{A}{B}=\frac{25}{1980}=\frac{5}{396}\)