\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\)

\(\dfrac{x-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-1+3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(1-3\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(-2\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\)

\(\left(x+2\right)\left(y+2\right)=9\)

=> (X+2) ; (y+2) ϵ Ư(9)

TH1: x+2 = 1 => x = -1

y+2=9 => y = 7

TH2: x+2 = 9 => x = 7

=> y +2 = 1 => y =-1

TH3:x+2 = -9 => x = -11

y+2 = -1 => y=-3

TH4: x+2 = -1 => x =-3

y+2 = -9 => x=-11

TH5: x+2 = -3 => x =-5

y+2 = -3 => y=-5

TH6: x+2 =3 =>  x = 1

y+2=3 => y=1

ĐKXĐ: x>=0; x<>1

PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)

=>6-2x=0

=>x=3

\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)

\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)

\(\Rightarrow2xy=54+y\)

\(\Rightarrow2xy-y=54\)

\(\Rightarrow xy-\dfrac{y}{2}=27\)

\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)

\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)

\(\left|x+\dfrac{1}{2}\right|+\left|x-y+z\right|+\left|y+\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y+\dfrac{1}{3}=0\\x-y+z=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{1}{3}\\z=-x+y=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\end{matrix}\right.\)

\(A=2x+y+z=-1-\dfrac{1}{3}+\dfrac{1}{6}=-\dfrac{4}{3}+\dfrac{1}{6}=-\dfrac{7}{6}\)

\(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

P>3/2

=>P-3/2>0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

=>-căn x+2>0

=>-căn x>-2

=>0<x<4

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

x = 2 hoặc x = -1