*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

\(\dfrac{3^{17}.81^{11}}{27^{10}.9^{15}}=\dfrac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{17}.3^{44}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

Vậy...............

Chúc bạn học tốt!!!

\(\dfrac{3^{17}.81^{11}}{27^{10}.9^{15}}=\dfrac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{17}.3^{44}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

theo bài ra ta có:

\(\dfrac{6}{x+1}.\dfrac{x-1}{3}=\dfrac{6x-6}{3x+1}\\ =\dfrac{6x+2-8}{3x+1}\\ =\dfrac{2\left(3x+1\right)-8}{3x+1}\\ =2-\dfrac{8}{3x+1}\)

để \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\) là số nguyên

=> \(\dfrac{8}{3x+1}\) nguyên

\(8⋮3x+1\\ \Rightarrow3x+1\inƯ_{\left(8\right)}=\left\{-1;1;2;-2;4;-4;8;-8\right\}\)

ta có bảng sau:

mà x là số nguyên

=> x ={0;-1;1;-3}

vậy x ={0;1;-1;-3}

giup mk voi mk dang can gap

      \(\frac{11x-2}{7x-5}=\frac{11}{8}\)

\(\Leftrightarrow8\left(11x-2\right)=11\left(7x-5\right)\)

\(\Leftrightarrow88x-16=77x-55\)

\(\Leftrightarrow88x-77x=-55+16\)

\(\Leftrightarrow11x=-39\)

\(\Leftrightarrow x=\frac{-39}{11}\)

\(\frac{12x-2}{7x-8}=\frac{11}{8}\)

\(\Rightarrow8\left(12x-2\right)=11\left(7x-8\right)\)

\(\Rightarrow96x-16=77x-88\)

\(\Rightarrow96x-77x=-88+16\)

\(\Rightarrow19x=-72\)

\(\Rightarrow x=-\frac{72}{19}\)

\(\overline{abc}+\overline{bca}+\overline{cab}⋮37\)

\(\Rightarrow1000.a+100.b+10.c⋮37\)

\(\Rightarrow1000a-999.a+100.b+10.c⋮37\)

\(\Rightarrow100.b+10.c+a=\overline{bca}⋮37\)

Thanks

\(A=\frac{x-2014}{x-14}=1-\frac{2000}{x-14}\)

A lớn nhất  khi x -14 lớn nhất 

=> không có x lớn nhất để x -14 lớn nhất

( đề sai )

Để A đạt đc GTLN 14 - x phải = 1

suy ra 14 - x = 1 khi x =13...............