Gọi số hàng đơn vị là a(\(a\in N;0\le a\le9\))

Ta có \(\overline{8a}=\left(8+a\right)^2\Rightarrow80+a=64+16a+a^2\)

<=> \(a^2+15a-16=0\)

<=> (a - 1)(a + 16) = 0

<=> \(\left[{}\begin{matrix}a=1\left(tm\right)\\a=-16\left(\text{loại}\right)\end{matrix}\right.\)

Vậy số cần tìm 81

1C; 2B ; 3C ; 4B ; 5A 

ĐKXĐ : \(x\ge-\dfrac{1}{3}\)

\(x^2+x-4\sqrt{3x+1}+6=0\)

<=> \(\left(x^2-2x+1\right)+\left(3x+1-4\sqrt{3x+1}+4\right)=0\)

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

<=> \(\left\{{}\begin{matrix}x-1=0\\\sqrt{3x+1}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3x+1=4\end{matrix}\right.\Leftrightarrow x=1\)

Vậy x = 1 là nghiệm phương trình 

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\left(a;b;c\ne0\right)\)

<=> ab + bc + ca = 1

Thay ab + bc + ca = 1 vào A ta được

 A = (a² + ab + bc + ca)(b² + ab + bc + ca)(c² + ab + bc + ca)

= (a + b)(a + c)(b + c)(a + b)(b + c)(c + a)

= [(a + b)(b + c)(c + a)]² 

=> A là bình phương của 1 số

Ta có : \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{2c+a}{9}+\dfrac{b}{3}\ge3\sqrt[3]{\dfrac{a^3}{b\left(2c+a\right)}.\dfrac{2c+a}{9}.\dfrac{b}{3}}=a\)

Tương tự ta được 

\(\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{2a+b}{9}+\dfrac{c}{3}\ge b\);

\(\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{a}{3}+\dfrac{2b+c}{9}\ge c\)

Cộng vế với vế 

=> \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2\left(a+b+c\right)}{3}\ge a+b+c\)

=> \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)

a) Ta có : \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

=> \(P:Q=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{-3}{\sqrt{x}+3}\)

Để P:Q < -1/2

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

<=> \(\dfrac{3}{\sqrt{x}+3}>\dfrac{1}{2}\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ : => Khi \(0\le x< 9\) thì P : Q < -1/2

b) Ta có \(\sqrt{x}+3\ge3\Rightarrow\dfrac{3}{\sqrt{x}+3}\le1\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\)

=> Min P:Q = -1

"=" khi x = 0

b) \(A=1\sqrt{4a+1}+1.\sqrt{4b+1}+1.\sqrt{4c+1}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4a+1+4b+1+4c+1\right)}\)

\(=\sqrt{3.\left[4\left(a+b+c\right)+3\right]}=\sqrt{21}\left(\text{vì }a+b+c=1\right)\)

"=" xảy ra <=> \(\dfrac{1}{\sqrt{4a+1}}=\dfrac{1}{\sqrt{4b+1}}=\dfrac{1}{\sqrt{4c+1}};a+b+c=1\)

<=> a = b = c = 1/3 

a) (x + y + z)² \(\le3\left(x^2+y^2+z^2\right)\)(1) 

<=> \(x^2+y^2+z^2+2xy+2yz+2zx\le3x^2+3y^2+3z^2\)

<=> \(2x^2+2y^2+2z^2-2xy-2xz-2yz\ge0\)

<=> (x - y)² + (y - z)² + (z - x)² \(\ge0\) (đúng) 

=> (1) đúng "=" khi x = y = z

Ta có \(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

=> \(\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2+\left(\dfrac{c}{z}\right)^2+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)(1)

Lại có \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

=> \(\dfrac{ayz+bxz+cxy}{abc}=0\Rightarrow\dfrac{yz}{bc}+\dfrac{xz}{ac}+\dfrac{xy}{ab}=0\)(2) 

Từ (1)(2) => ĐPCM

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=6\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y}=7\\\dfrac{1}{x}+\dfrac{1}{y}=2\end{matrix}\right.\)

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

Vậy (x;y) = (1;1) là nghiệm hệ phương trình