\(\frac{2x+1}{3}=\frac{x-5}{2}\)

\(\Rightarrow2\left(2x+1\right)=3\left(x-5\right)\)

\(\Rightarrow4x+2=3x-15\)

\(\Rightarrow4x-3x=-15-2\)

\(\Rightarrow x=-17\)

Ta có:

\(\frac{2x+1}{3}=\frac{x-5}{2}\)

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

\(\Rightarrow4x+2=3x-15\)

\(\Rightarrow4x-3x=-15-2\)

\(\Rightarrow x=-17\)

Vậy x = -17

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)}

\(A=\dfrac{2x+1+4}{2x+1}=1+\dfrac{4}{2x+1}\)

A min khi 2x+1=-1

=>x=-1

-2x - (x - 7) = 34 - (-x + 25)

-3x+7=34+x-25

-4x=2

x=-1/2

-2x-(x-7)=34-(-x+25)

-2x-x+7=34+x-25

-2x-x+7-34-x+25=0

-4x-2=0

-4x=2

x=\(-\frac{1}{2}\)

#H

(3k + 1)² = (3k + 1).(3k + 1) = 3k.(3k + 1) + 1.(3k + 1)

              = 9k² + 3k + 3k + 1

              = 9k² + 6k + 1