a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

\(pt\Leftrightarrow\left|2x-3\right|+2\left|2x-3\right|=3\)\(\Leftrightarrow3\left|2x-3\right|=3\Leftrightarrow\left|2x-3\right|=1\)

\(\Leftrightarrow2x-3=1\text{ hoặc }2x-3=-1\)

\(\Leftrightarrow x=2\text{ hoặc }x=1\)

\(\text{Vậy, }x\in\left\{2;1\right\}\)

tk cho mình đi mãi yêu

Ta có: 2(x+1)=3(4x-1)

=>12x-3=2x+2

=>10x=5

hay x=1/2

Thay x=1/2 vào P, ta được:

\(P=\dfrac{2\cdot\dfrac{1}{2}+1}{2\cdot\dfrac{1}{2}+5}=\dfrac{1+1}{1+5}=\dfrac{2}{6}=\dfrac{1}{3}\)

1: \(\frac{5x+2}{5}>4x+3\)

=>5x+2>5(4x+3)

=>5x+2>20x+15

=>-15x>13

=>\(x<-\frac{13}{15}\) (2)

\(\frac{8x+3}{3}<2x+7\)

=>8x+3<3(2x+7)

=>8x+3<6x+21

=>2x<18

=>x<9(1)

Từ (1),(2) suy ra \(x<-\frac{13}{15}\)

mà x là số nguyên dương

nên x∈∅

2: \(\frac{x-1}{3}-\frac{2-x}{4}\ge\frac13\)

=>\(\frac{4\left(x-1\right)-3\left(2-x\right)}{12}\ge\frac{4}{12}\)

=>4(x-1)-3(2-x)>=4

=>4x-4-6+3x>=4

=>7x-10>=4

=>7x>=14

=>x>=2(3)

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

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

=>x>=-5(4)

Từ (3),(4) suy ra x>=2

mà x là số nguyên dương

nên x∈{2;3;4;5;6;...}

Đặt $a=2x,\quad b=3y,\quad c=4z.$

Khi đó $a^2+b^2+c^2=1.$

Ta có $A=\dfrac{a}{1-a^2}+\dfrac{b}{1-b^2}+\dfrac{c}{1-c^2}.$

Xét hàm số $f(t)=\dfrac{t}{1-t^2}\quad(0<t<1).$

Ta có $f''(t)=\dfrac{2t(3+t^2)}{(1-t^2)^3}>0.$

Do đó $f(t)$ là hàm lồi trên $(0;1)$.

Theo bất đẳng thức Jensen:

$A\ge3f\left(\dfrac{a+b+c}{3}\right).$

Mặt khác, theo bất đẳng thức Cauchy:

$(a+b+c)^2\le3(a^2+b^2+c^2)=3$

$\Rightarrow\dfrac{a+b+c}{3}\le\dfrac1{\sqrt3}.$

Vì $f(t)$ đồng biến trên $(0;1)$ nên

$A\ge3f\left(\dfrac{a+b+c}{3}\right)$

$\ge3f\left(\dfrac1{\sqrt3}\right)$

$=3\cdot\dfrac{\frac1{\sqrt3}}{1-\frac13}$

$=3\cdot\dfrac{\frac1{\sqrt3}}{\frac23}$

$=\dfrac{3\sqrt3}{2}.$

$a=b=c=\dfrac1{\sqrt3}.$

Dấu '=' xảy ra khi:

\Leftrightarrow x=\dfrac1{2\sqrt3},\quady=\dfrac1{3\sqrt3},\quadz=\dfrac1{4\sqrt3}.

$A_{\min}=\dfrac{3\sqrt3}{2}.$