\(\Leftrightarrow\)2(9x²+6x+1)=(3x+1)(x-2)

\(\Leftrightarrow\)2(3x+1)²-(3x+1)(x-2)=0

\(\Leftrightarrow\)(3x+1)[2(3x+1)-(x-2)]=0

\(\Leftrightarrow\)(3x+1)(6x+2-x+2)=0

\(\Leftrightarrow\)(3x+1)(5x+4)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\Leftrightarrow3x=-1\Leftrightarrow x=\frac{-1}{3}\\5x+4=0\Leftrightarrow5x=-4\Leftrightarrow x=\frac{-4}{5}\end{cases}}\)​

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)

Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.

\(B=\frac{1}{2020}\)

B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)

    = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)

    = \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)

   =  \(\frac{1}{2020}\)

 +)\(\frac{x}{3}=0\)  \(\Rightarrow x=0\)

+) \(\frac{x}{3}< 0\)  \(\Rightarrow x< 0\)

+)\(1< \frac{x}{3}< 2\)    \(\Rightarrow3< x< 6\)

x/3=0 thì x=0

1: 6(x-2)-y(2-x)=10

=>6(x-2)+y(x-2)=10

=>(x-2)(y+6)=10

=>(x-2;y+6)∈{(1;10);(10;1);(-1;-10);(-10;-1);(2;5);(5;2);(-2;-5);(-5;-2)}

=>(x;y)∈{(3;4);(12;-5);(1;-16);(-8;-7);(4;-1);(7;-4);(0;-11);(-3;-8)}

2: 3x-2xy+3y=6

=>x(3-2y)+3y-4,5=6-4,5

=>-x(2y-3)+1,5(2y-3)=1,5

=>(2y-3)(-x+1,5)=1,5

=>(2y-3)(-2x+3)=3

=>(2x-3)(2y-3)=-3

=>(2x-3;2y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(2;0);(0;2);(1;3);(3;1)}

3: 6x-xy+2y=5

=>x(6-y)+2y-12=5-12=-7

=>-x(y-6)+2(y-6)=-7

=>(y-6)(-x+2)=-7

=>(x-2)(y-6)=7

=>(x-2;y-6)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(x;y)∈{(3;13);(9;7);(1;-1);(-5;5)}

\(\left(3x+1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)

\(\Leftrightarrow3x^2+10x+3=10-11x+3x^2\)

\(\Leftrightarrow21x=7\)

\(\Leftrightarrow x=\frac{7}{21}=\frac{1}{3}\)

Vậy : \(x=\frac{1}{3}\)

a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)

\(=64-x^3+\left(x-4\right)^3\)

\(=64-x^3+x^3-12x^2+48x-64\)

\(=-12x^2+48x\)

b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(=27x^3+8-27x^3+8\)

=16

c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)

\(=x^3+1-x\left(x^2+2x+1\right)\)

\(=x^3+1-x^3-2x^2-x\)

\(=-2x^2-x+1\)

a) 5^x + 5^x+2 = 650

   5^x + 5^x+2 = 5⁴ + 5²

   5^x +  5^x+2 = 5⁶

 ^{\(\Rightarrow\)}x+x+2 = 6

         2x       = 6-2 

          x        = 2

b) 3x - ( 2x +1) = 2

   3x - 2x -1     = 2 

    3x - 2x         =2+1 

   (3 - 2).x         = 3

     1.x             = 3

       x              = 3

sao ko k  cho mình ?

GTNN là -4 khi x=1