Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)

\(\Leftrightarrow\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)

\(\Leftrightarrow\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}\)

mà z-x=-196

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}=\frac{18z-18x}{5-33}=\frac{18\left(z-x\right)}{-28}=\frac{-18\cdot196}{-28}=126\)

Do đó:

\(\left\{{}\begin{matrix}\frac{6x}{11}=126\\\frac{9}{2}y=126\\\frac{18z}{5}=126\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=1386\\y=28\\18z=630\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=231\\y=28\\z=35\end{matrix}\right.\)

Vậy: (x,y,z)=(231;28;35)

Ta có : \(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)

=> \(\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)=\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)\)

=> \(\frac{x+2+198}{198}+\frac{x+3+197}{197}=\frac{x+4+196}{196}+\frac{x+5+195}{195}\)

=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)

=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)

=> \(\left(x+200\right)\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)=0\)

Ta có : \(\frac{1}{198}+\frac{1}{197}\ne\frac{1}{196}+\frac{1}{195}\)  =>    \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\ne0\)

=> x + 200 = 0

=> x = -200

<=> (\(\frac{x+2}{198}\)+1) +(\(\frac{x+3}{197}\)+1) =(\(\frac{x+4}{196}\)+1) +(\(\frac{x+5}{195}\)+1)

<=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)

<=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)

<=> \(\left(x+200\right)\cdot\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)\)=0

Vì \(\frac{1}{195}>\frac{1}{196}>\frac{1}{197}>\frac{1}{198}\)

<=> \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\) khác 0

<=> \(x+200=0\)

<=> x       = 

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

\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{-x+z}{-\dfrac{11}{6}+\dfrac{5}{18}}=\dfrac{-196}{-\dfrac{14}{9}}=126\)

Do đó: x=231; y=28; z=35

Lời giải:

Ta có:

\(\frac{6}{11}x=\frac{6x}{11}=\frac{18x}{33}\)

\(\frac{9}{2}y=\frac{9y}{2}=\frac{18y}{4}\)

Mà: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) => \(\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}\)

Theo đề bài, ta có: y - x + z = -196

=> Áp dụng tính chất của dãy tỉ số bằng nhau, ta có

\(\frac{18y}{4}=\frac{18x}{33}=\frac{18z}{5}=\frac{18y-18x+18z}{4-33+5}=\frac{18\left(y-x+z\right)}{-24}=\frac{-18.196}{-24}=\frac{3528}{24}=147\)

=>\(\left\{{}\begin{matrix}\frac{6}{11}x=147\Leftrightarrow x=147.\frac{11}{6}=\frac{539}{2}\\\frac{9}{2}y=147\Leftrightarrow y=147.\frac{2}{9}=\frac{98}{3}\\\frac{18}{5}z=147\Leftrightarrow z=147.\frac{5}{18}=\frac{245}{6}\end{matrix}\right.\) (TMĐK)

Vậy: \(x=\frac{539}{2};y=\frac{98}{3};z=\frac{245}{6}\)

Chúc bạn học tốt!Tick cho mình nhé!

\(a)\) \(M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}\)

\(3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}\)

\(3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)

\(2M_{\left(3\right)}=3^{2017}-3\)

\(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)

Vậy \(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)

\(M_{\left(-3\right)}=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2016}\)

\(\left(-3\right)M_{\left(-3\right)}=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+...+\left(-3\right)^{2017}\)

\(\left(-3\right)M_{\left(-3\right)}-M_{\left(-3\right)}=\left[\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2017}\right]-\left[\left(-3\right)+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]\)\(\left(-4\right)M_{\left(-3\right)}=\left(-3\right)^{2017}+3\)

\(M_{\left(-3\right)}=\frac{\left(-3\right)^{2017}+3}{-4}\)

\(M_{\left(-3\right)}=\frac{-\left(3^{2017}-3\right)}{-4}\)

\(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)

Vậy \(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)

Chúc bạn học tốt ~ 

\(b)\) Ta có : 

\(M_{\left(2\right)}=2+2^2+2^3+...+2^{2016}\)

\(M_{\left(2\right)}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)

\(M_{\left(2\right)}=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)\)

\(M_{\left(2\right)}=2.7+2^4.7+...+2^{2014}.7\)

\(M_{\left(2\right)}=7\left(2+2^4+...+2^{2014}\right)⋮7\) \(\left(1\right)\)

Lại có : 

\(M_{\left(2\right)}=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)

\(M_{\left(2\right)}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2013}\left(1+2+2^2+2^3\right)\)

\(M_{\left(2\right)}=2.15+2^5.15+...+2^{2013}.15\)

\(M_{\left(2\right)}=15\left(2+2^5+...+2^{2013}\right)⋮15\) \(\left(2\right)\)

Từ (1) và (2) suy ra \(M_{\left(2\right)}\) chia hết cho \(7\) và \(15\)

\(\Rightarrow\)\(M_{\left(2\right)}⋮105\) ( vì \(7.15=105\) ) 

Vậy nếu \(M⋮105\)\(\Leftrightarrow\)\(x=2\)

Chúc bạn học tốt ~ 

a, \(M=7.\left(x-y\right)+4a.\left(x-y\right)-5\)

Theo bài ra ta có: x-y=0

=> \(M=0+0-5\)

\(\Rightarrow M=-5\)

b,

\(N=\left(x^2+y^2\right).\left(x-y\right)+3\)

\(\Rightarrow N=0+3=3\)

lớp 7 lên 8 à làm quen nhá :)

a) \(M=7x-7y+4ax-4ay-5\)

\(M=7\left(x-y\right)+4a\left(x-y\right)-5\)

\(M=0+0-5=-5\)

b) \(N=x\left(x^2+y^2\right)-y\left(x^2+y^2\right)+3\)

\(N=\left(x-y\right)\left(x^2+y^2\right)+3\)

\(N=0+3=3\)

\(A.\dfrac{-5}{13}.\dfrac{-4}{13}.\dfrac{-3}{13}.....\dfrac{4}{13}.\dfrac{5}{13}\)

\(A=\dfrac{-5.-4.-3.-2.-1.0.1.2.3.4}{13^{10}}\)

\(A=\dfrac{0}{13^{10}}=0\)

\(A=\dfrac{-5}{13}\cdot\dfrac{-4}{13}\cdot\dfrac{-3}{13}\cdot...\cdot\dfrac{4}{13}\cdot\dfrac{5}{13}\)

\(A=\dfrac{\left(-5\cdot5\right)\cdot\left(-4\cdot4\right)\cdot...\cdot\left(-1\cdot1\right)\cdot0}{13\cdot13\cdot13\cdot...\cdot13}\)

\(A=\dfrac{0}{13\cdot13\cdot13\cdot...\cdot13}\)

\(A=0\)

\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+z}{-\frac{11}{6}+\frac{5}{18}}=-\frac{196}{-\frac{14}{9}}=126\)

x=\(126.\frac{11}{6}=231\)

y=\(126.\frac{2}{9}=28\)

z=\(126.\frac{5}{18}=35\)

b) \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{96}{19}\)hoặc \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=-\frac{96}{19}\)

=> ...........