Ôn tập chương 1

3x=4y=> \(\dfrac{x}{3}=\)\(\dfrac{y}{4}\)=>\(\dfrac{x^2}{3^2}=\) \(\dfrac{y^2}{4^2}\)

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{3^2}=\)\(\dfrac{y^2}{4^2}\)=\(\dfrac{x^2+y^2}{9+16}=\)\(\dfrac{400}{25}=\)16

+) \(\dfrac{x^2}{9}=16\Rightarrow x^2=9.16=144\Rightarrow x=\pm12\)

+) \(\dfrac{y^2}{16}=16\Rightarrow y^2=16.16=256\Rightarrow y=\pm16\)

vẬY X=12 ,y=16 hoặc x=-12,y=-16

a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)

\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2011}-2^0\)

\(\Rightarrow A=2^{2011}-1\)

b,\(7^{x+2}+2.7^{x-1}=345\)

\(7^{x-1}.\left(7^3+2\right)=345\)

\(\Rightarrow7^{x-1}.345=345\)

\(\Rightarrow7^{x-1}=345:345=1\)

\(\Rightarrow7^{x-1}=7^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)

\(A=\dfrac{101+102+103+...+109}{100}\)

Xét tử số : \(101+102+103+...+199\)

Có : \(\left(199-101\right):1+1=99\) (số hạng)

\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)

\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)

\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

Xét tử số : \(10002+10004+10006+...+12014\)

Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)

\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)

\(\Rightarrow B=\dfrac{11085056}{10000}\)

Bạn tự làm câu C nha

\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)

\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)

\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)

\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)

\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)

D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)

E = \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}\)+ ...+ \(\dfrac{1}{2014.2017}\)
E = \(\dfrac{1.3}{3.1.4}+\dfrac{1.3}{3.4.7}+\dfrac{1.3}{3.7.10}+...+\dfrac{1.3}{3.2014.2017}\)
E = \(\dfrac{1}{3}\) .( \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\) )
E = \(\dfrac{1}{3}\).\(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(1-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(\dfrac{2015}{2015}-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\dfrac{2014}{2015}\)
E = \(\dfrac{2014}{6045}\)

\(2x=4z\Rightarrow z=\dfrac{x}{2}\)

\(2x=-3y\Rightarrow y=\dfrac{-2}{3}x\)

Thay vào \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{\dfrac{-2}{3}x}+\dfrac{1}{\dfrac{x}{2}}=3\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{\dfrac{-2}{3}.\dfrac{-3}{2}.x}+\dfrac{2}{2\dfrac{x}{2}}=3\)

\(\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{x}+\dfrac{2}{x}\)

\(\Rightarrow\dfrac{\left(1+\dfrac{-3}{2}+2\right)}{x}=3\)

\(\Rightarrow\dfrac{\dfrac{3}{2}}{x}=3\)

\(\Rightarrow x=\dfrac{1}{2}\)

\(z=\dfrac{x}{2}=\dfrac{\dfrac{1}{2}}{2}=\dfrac{1}{4}\)

\(y=\dfrac{-2}{3}x=\dfrac{-2}{3}.\dfrac{1}{4}=\dfrac{-1}{6}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{6}\end{matrix}\right.\)

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)

\(=\dfrac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)

0,(72) =8 phần 11

0.2 (36) = 13 phần 55

3,(234)=359 phần 111

2,71 (3)= 407 phần 150

Tick đúng nha

0,944 - 2x = 3,268

=> 2x = 3,268 + 0.944

=> 2x = 4,212

=> x = 4,212 : 2 = 2,106

0,944 - 2x = 3.268

0,944 - 2x = 804

2x = 804 + 0,944

2x = 804,944

x = 804,944 : 2

x = 402,472

a) \(9\cdot3^2\cdot\dfrac{1}{81}=\dfrac{9\cdot9}{81}=\dfrac{81}{81}=1\)

b) \(2\dfrac{1}{2}+\dfrac{4}{7}:\left(-\dfrac{8}{9}\right)\)\(=\dfrac{5}{2}+\dfrac{4}{7}\cdot\left(-\dfrac{9}{8}\right)\)

\(=\dfrac{5}{2}-\dfrac{9}{14}\)\(=\dfrac{35}{14}-\dfrac{9}{14}=\dfrac{26}{14}=\dfrac{13}{7}\)

c) \(3,75\cdot7,2+2,8\cdot3,75=3,75\left(7,2+2,8\right)=3,75\cdot10=37,5\)

Ta có :

\(5.2^{x-1}-2^x=96\\ \Rightarrow5.2^x:2-2^x=96\\ \Rightarrow2^x.\left(5:2-1\right)=96\\ \Rightarrow2^x.\dfrac{3}{2}=96\\ \Rightarrow2^x=64\\ \Rightarrow2^x=2^6\\ \Rightarrow x=6\)

Vậy x=6