\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x+5}}=\dfrac{2\sqrt{x}+10}{\sqrt{x}+5}+\dfrac{7}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\) 

Để \(A\) ∈ \(Z\) thì \(\dfrac{7}{\sqrt{x}+5}\) phải ∈ \(Z\)

=> \(\sqrt{x}+5\) ∈ \(Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)

# Với \(\sqrt{x}+5=-7=>\sqrt{x}=-12\)(Loại)

#Với \(\sqrt{x}+5=-1=>\sqrt{x}=-6\)(Loại)

#Với \(\sqrt{x}+5=1=>\sqrt{x}=-4\left(Loại\right)\)

#Với \(\sqrt{x}+5=7=>\sqrt{x}=2< =>x=4\left(Nhận\right)\)

Vậy \(x=4\) thì \(A\)∈\(Z\)

\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}3\)

\(Ta\) \(Có\) : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}=\sqrt[3]{\dfrac{a^6}{ab.ab\left(a^2-ab+b^2\right)}}=\dfrac{a^2}{\sqrt[3]{ab.ab.\left(a^2-ab+b^2\right)}}\) 

\(Áp\) \(dụng\) \(bđt\) \(AM-GM\) 

\(\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}\text{≤}\)  \(\dfrac{ab+ab+a^2-ab+b^2}{3}\) 

\(=>\dfrac{a^2}{\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}}\) \(\text{≥}\) \(\dfrac{3a^2}{a^2+ab+b^2}\) \(Hay\) \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}\text{≥}\dfrac{3a^2}{a^2+ab+b^2}\)

Tương tự ta cũng có : 

\(\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\text{≥}\dfrac{3b^2}{b^2+bc+c^2}\) 

\(\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+a^2\right)}}\text{≥}\dfrac{3c^2}{a^2+ac+c^2}\)

\(=>\text{​​}\text{​​}\)\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\)  \(\text{≥}\) \(3\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\) 

Cần c/m \(\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\) ≥ \(1\) 

Ta có : \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\) 

\(< =>3a^2\text{≥}a^2+ab+b^2\) \(< =>2a^2-b\left(a+b\right)\text{≥}0\) (1)

Lại có : \(a^2\text{≥}-b\left(a+b\right)\) (2)

Từ (1) và (2) => \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)

Tương tự ta cũng có :

 \(\dfrac{b^2}{b^2+bc+c^2}\text{≥}\dfrac{1}{3}\) 

\(\dfrac{c^2}{a^2+ac+c^2}\text{≥}\dfrac{1}{3}\)

Do đó \(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\text{≥}1\)

Suy ra :  \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\) 

Đẳng thức xảy ra <=> \(a=b=c=1\)

1.2 với \(x\ge0,x\in Z\)

A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)

*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)

*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)

*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)

vậy x=1 thì A\(\in Z\)

Bài 1:

\(a)\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}.\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3.\\ \Rightarrow x+\dfrac{2}{3}=\dfrac{5}{4}.\\ \Leftrightarrow x=\dfrac{7}{12}.\)

\(b)\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}.\\\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right) ^3.\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{2}{7}.\\ \Leftrightarrow x=\dfrac{11}{14}.\)

Bài 2:

\(a)\left(x-\dfrac{1}{3}\right)^2=\dfrac{25}{9}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{5}{3}\right)^2.\\\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{-5}{3}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{5}{3}.\\x-\dfrac{1}{3}=\dfrac{-5}{3}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-4}{3}.\end{matrix}\right.\)

\(b)\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{7}{4}\right)^2.\\\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{-7}{4}\right)^2.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}.\\x-\dfrac{3}{4}=\dfrac{-7}{4}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}.\\x=-1.\end{matrix}\right.\)

bị lỗi nhé

= \(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)

= \(\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\)

=10-\(\dfrac{32}{5}\)

=\(\dfrac{18}{5}\)

\(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)                                                                                                         \(=\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\) 

\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)

\(=10-\dfrac{32}{5}\)

\(=\dfrac{50}{5}-\dfrac{32}{5}\)

\(=\dfrac{18}{5}\)

15:

a: \(\text{Δ}=\left(m^2-m+2\right)^2-4m^2\)

=(m^2-m+2-2m)(m^2-m+2+2m)

=(m^2+m+2)(m^2-3m+2)

=(m-1)(m-2)(m^2+m+2)

Để phương trình co hai nghiệm phân biệt thì (m-1)(m-2)(m^2+m+2)>0

=>(m-1)(m-2)>0

=>m>2 hoặc m<1

b: x1+x2=m^2-m+2>0 với mọi m

x1*x2=m^2>0 vơi mọi m

=>Phương trình luôn có hai nghiệm dương phân biệt

Bách chiến bách thắng

Kim chi ngọc diệp