Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A= 1+2-3-4+5+6-7-8+...+2013+2014
A=(1+2-3-4)+(5+6-7-8)+.....+(2013+2014)
A=(-4)+(-4)+...+(-4)+4027
A=(-4).503+4027
A=-2012+4027
A=2015
B=\(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2016}{2015}\)
B=\(\dfrac{3.4.5.6.....2016}{2.3.4.5.....2015}=\dfrac{2016}{2}=1008\)
1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)
2) thiếu quế phải nha
3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)
\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)
\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)
4) ta có : \(x+y=20\Leftrightarrow y=20-x\)
\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)
\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)
\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)
\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)
\(\dfrac{23+x}{40-x}=\dfrac{-3}{4}\Leftrightarrow4\left(23+x\right)=-3\left(40-x\right)\)
\(\Leftrightarrow92+4x=-120+3x\Leftrightarrow4x-3x=-120-92\)
\(\Leftrightarrow x=-212\) vậy \(x=-212\)
Ta có:
\(A=\dfrac{9}{a^{2013}}+\dfrac{7}{a^{2014}}\)
\(=\left(\dfrac{8}{a^{2013}}+\dfrac{1}{a^{2013}}\right)+\left(\dfrac{8}{a^{2014}}-\dfrac{1}{a^{2014}}\right)\)
\(=\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)\)
\(B=\dfrac{8}{a^{2014}}+\dfrac{8}{a^{2013}}\)
\(=\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)
Vì \(\dfrac{1}{a^{2013}}>\dfrac{1}{a^{2014}}\Rightarrow\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}>0\)
\(\Rightarrow\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)>\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)
Vậy \(A>B\)
1. x3 - \(\dfrac{4}{25}\)x = 0
<=> x(x2 - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh
1 ) \(x^3-\dfrac{4}{25}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)
Vậy .............
2 ) \(3^{4x+4}=9^{x+2}\)
\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)
\(\Leftrightarrow4x+4=2x+4\)
\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)
3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )
4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)
\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)
Vậy ......
5) \(2+4+6+...+2x=210\)
\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)
\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)
\(\Leftrightarrow1+2+3+...+x=105\)
\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)
\(\Leftrightarrow x\left(x+1\right)=210\)
Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)
\(\Leftrightarrow x=14\)
Vậy ......
6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x=17.\)
Vậy ...........
\(\)
Hướng dẫn + lời giải
\(A=-1-2+3+4+..-2013-2014+2015+2016\)
A có 2016 số hạng
quy luật (2 trừ đến 2 cộng)
A chia hết cho 4 =>ghép 4 số hạng
\(B=\left(-1-2+3+4\right)+\left(-5-6+7+8\right)+...+\left(-2013-2014+2015+2016\right)\\ \)
\(C=4+4+4+...+4\)
số số hạng của C số số hạng của A chia 4
\(\dfrac{2016}{4}=504\)
Vậy C=4.504=2016
mình cố tình đặt A,B,C để bạn dẽ hiểu bản chất nó vẫn là A
bài có n! cách làm
cách này hứơng bạn đi đến cái tổng quát --> có thể làm được toán lớp 11
5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)
\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)
\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)
\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)
1) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-\left(x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-x=2+4\\5x+x=-2+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{3}{2}\)
2) \(\left|x+15\right|=\left|3x-4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-\left(3x-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-3x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-4-15\\x+3x=4-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-19\\4x=-11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
Vậy \(x=-\dfrac{11}{4}\) hoặc \(x=\dfrac{19}{2}\)
3) \(\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|-\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|=\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\left(\dfrac{5}{8}x+\dfrac{3}{5}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{8}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy \(x=\dfrac{116}{75}\) hoặc \(x=\dfrac{164}{25}\)
4) \(\left|2x-6\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|2x-6\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-\left(x+3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3+6\\2x+x=-3+6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\3x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=9\)
1: \(A=x^2+y^2+2014\ge2014\)
Dấu '=' xảy ra khi x=y=0
2: \(B=\left(x+30\right)^2+\left(y-4\right)^2+17\ge17\)
Dấu '=' xảy ra khi x=-30 và y=4
3: \(C=\left(y-9\right)^2+\left|x-3\right|-1\ge-1\)
Dấu '=' xảy ra khi x=3 và y=9
Thế bạn có làm được không Võ Nguyễn Anh Thư? Trả lời thì trả lời câu hỏi ý, trả lời cái đấy để làm gì?
Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!