b,2x.(x-5)-x.(3+2x)=26

2x² - 10x - 3x - 2x² = 26

-13x = 26

x = -2

c, (x+7)²-x.(x-3)=12

x² +14x +49 - x² + 3x = 12

17x + 49 = 12

17x = - 37

x = \(\dfrac{-37}{17}\)

d, 9( x -2018) - x+ 2018 =0

9( x -2018) - (x -2018) = 0

( 9-1)(x -2018) = 0

8( x -2018) = 0

x -2018 = 0

x = 2018

a: =>2x+10-x^2-5=0

=>-x^2+2x+5=0

=>\(x\in\left\{1+\sqrt{6};1-\sqrt{6}\right\}\)

e: =>4x^2+4x+9x^2-4=15

=>13x^2+4x-19=0

=>\(x\in\left\{\dfrac{-2+\sqrt{251}}{13};\dfrac{-2-\sqrt{251}}{13}\right\}\)

0,2083.................

\(A=\left(26^{2018}+3^{2018}\right)^{2019}\)

\(B=\left(26^{2019}+3^{2019}\right)^{2018}\)

\(B=\left(26^{2018}.26+3.3^{2018}\right)^{2018}< \left(26^{2018}.26+3^{2018}.26\right)^{2018}\)

\(B< \left(26^{2018}+3^{2018}\right)^{2018}.26^{2018}< \left(26^{2018}+3^{2018}\right)^{2018}.\left(26^{2018}+3^{2018}\right)\)

\(\Rightarrow B< \left(26^{2018}+3^{2018}\right)^{2019}\Rightarrow B< A\)

Ta có: \(\frac{2x-4y}{39}=\frac{4z-3x}{26}=\frac{3y-2z}{52}\)

\(\Rightarrow\frac{39\left(2x-4y\right)}{39.39}=\frac{26\left(4z-3x\right)}{26.26}=\frac{52\left(3y-2z\right)}{52.52}\)

\(\Rightarrow\frac{78x-156y}{1521}=\frac{104z-78x}{676}=\frac{156y-104z}{2704}\)

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

\(\frac{78x-156y}{1521}=\frac{104z-78x}{676}=\frac{156y-104z}{2704}=\frac{78x-156y+104z-78x+156y-104z}{1521+676+2704}=\frac{0}{4901}=0\)

Do đó: \(\hept{\begin{cases}\frac{2x-4y}{39}=0\\\frac{4z-3x}{26}=0\\\frac{3y-2z}{52}=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-4y=0\\4z-3x=0\\3y-2z=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=4y\\4z=3x\\3y=2z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{4}=\frac{y}{2}\\\frac{z}{3}=\frac{x}{4}\\\frac{y}{2}=\frac{z}{3}\end{cases}}\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

Đặt \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}=k\)\(\Rightarrow\hept{\begin{cases}x=4k\\y=2k\\z=3k\end{cases}}\)

Ta có: \(A=2018-2x-11y+10z=2018-2.4k-11.2k+10.3k=2018-8k-22k+30k\)

\(A=2018-\left(8k+22k-30k\right)=2018-0=2018\)

chịu ?_?

\(\frac{2x-4y}{39}=\frac{4z-3x}{26}=\frac{3y-2z}{52}\)

\(\frac{2x-4y}{13\cdot3}=\frac{4z-3x}{13\cdot2}=\frac{3y-2z}{13\cdot4}\)=>\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}\) =\(\frac{3\left(2x-4y\right)}{3^2}=\frac{2\left(4z-3x\right)}{2^2}=\frac{4\left(3y-2z\right)}{4^2}\)=\(\frac{6x-12y}{9}=\frac{8z-6x}{2}=\frac{12y-8z}{4}\)

\(\text{Áp dụng tính chất của dãy tỉ số bằng nhau có:}\)

\(\frac{6x-12y}{9}=\frac{8z-6x}{2}=\frac{12y-8z}{4}\)=\(\frac{6x-12y+8z-6x+12y-8z}{9+2+4}\)=\(\frac{0}{15}\)=\(0\)

\(\Rightarrow\hept{\begin{cases}\frac{6x-12y}{9}=0\\\frac{8z-6x}{4}=0\\\frac{12y-8z}{16}=0\end{cases}}\hept{\begin{cases}6x-12y=0\\8z-6x=0\\12y-8z=0\end{cases}}\Rightarrow\)\(\hept{\begin{cases}6x=12y\\8z=6x\\12y=8z\end{cases}}\hept{\begin{cases}\frac{x}{12}=\frac{y}{6}\\\frac{z}{6}=\frac{x}{8}\\\frac{y}{8}=\frac{z}{12}\end{cases}}\Rightarrow\)\(\hept{\begin{cases}\frac{x}{4}=\frac{y}{2}\\\frac{z}{3}=\frac{x}{4}\\\frac{y}{2}=\frac{z}{3}\end{cases}}\)\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

Đặt \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}=k\)\(\Rightarrow\hept{\begin{cases}x=4k\\y=2k\\z=3k\end{cases}}\)

\(\Rightarrow A=\)\(2018-2\cdot4k-11\cdot2k+10\cdot3k\)\(=2018-8k-22k+30k=2018-0=2018\)

Ta có: \(\left(26^{2018}+3^{2018}\right)^{2019}=26^{2018\cdot2019}+3^{2018\cdot2019}\left(1\right)\)

          \(\left(26^{2019}+3^{2019}\right)^{2018}=26^{2019\cdot2018}+3^{2019\cdot2018}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left(26^{2018}+3^{2018}\right)^{2019}=\left(26^{2019}+3^{2019}\right)^{2018}\)