Ta có \(3x^2+4y^2+6x+3y-4=0\Rightarrow3x^2+6x+3+4\left(y^2+\dfrac{3}{4}y-\dfrac{7}{4}\right)=0\Rightarrow3\left(x+1\right)^2+4\left[\left(y+\dfrac{3}{8}\right)^2-\dfrac{121}{64}\right]=0\Rightarrow3\left(x+1\right)^2+\left(2y+\dfrac{3}{4}\right)^2=\dfrac{121}{16}\Rightarrow3\left(x+1\right)^2=\dfrac{121}{16}-\left(2y+\dfrac{3}{4}\right)^2\le\dfrac{121}{16}\Rightarrow\left(x+1\right)^2\le\dfrac{121}{48}< 3\)Do đó \(\left(x+1\right)^2\in\left\{0,1\right\}\)

-Xét \(\left(x+1\right)^2=0\Rightarrow x=-1\).Thay vào, ta có y = 1.

-Xét \(\left(x+1\right)^2=1\Rightarrow x+1\in\left\{1,-1\right\}\Rightarrow x\in\left\{0,-2\right\}\).Thay vào, ta có y ∈ ∅

Ta có \(4x^2+4x+y^2-6y=24\Rightarrow\left(2x+1\right)^2+\left(y-3\right)^2=34\)

Ta có \(\left(2x+1\right)^2\) là số chính phương lẻ, \(\left(y-3\right)^2\) là số chính phương.

Do đó \(\left(2x+1\right)^2\in\left\{1,9,25\right\}\)

-Xét \(\left(2x+1\right)^2=1\Rightarrow\left(y-3\right)^2=33\) (loại)

-Xét \(\left(2x+1\right)^2=9\Rightarrow\left\{{}\begin{matrix}\left(y-3\right)^2=25\\2x+1\in\left\{3,-3\right\}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y\in\left\{8,-2\right\}\\x\in\left\{1,-2\right\}\end{matrix}\right.\)

Với TH \(\left(2x+1\right)^2=25\) thì bạn làm tương tự là được nhá.

Ta có \(\dfrac{4}{x}-\dfrac{1}{y}=\dfrac{1}{2}\Rightarrow\dfrac{4y-x}{xy}=\dfrac{1}{2}\Rightarrow8y-2x=xy\Rightarrow xy+2x-8y-16=-16\Rightarrow\left(x-8\right)\left(y+2\right)=-16\)

Vì x,y nguyên nên đến đây bạn lập bảng để tim x,y nhá.

Vì \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\Rightarrow ca^2+b^2c=ab^2+c^2a\Rightarrow ca^2+b^{2c}-ab^2-c^2a=0\Rightarrow ca\left(a-c\right)-b^2\left(a-c\right)=0\Rightarrow\left(a-c\right)\left(ca-b^2\right)=0\)Mà \(a\ne c\) => \(ca=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\).

Khi đó, ta được \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{b}{c}\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}=\dfrac{10a+b-c}{10b+c-c}=\dfrac{a}{b}\) (luôn đúng)

=> đpcm

a)Vì \(\dfrac{n^4-n^2-1}{n+1}=\dfrac{n^2\left(n^2-1\right)-1}{n+1}=\dfrac{n^2\left(n-1\right)\left(n+1\right)-1}{n+1}=n^2\left(n-1\right)-\dfrac{1}{n+1}\)là số nguyên => \(\dfrac{1}{n+1}\) là số nguyên => \(n+1\in\left\{1,-1\right\}\) => \(n\in\left\{0.-2\right\}\) (thỏa mãn)

b)Vì \(\dfrac{n^3+2n-1}{n-2}=\dfrac{n^3-2n^2+2n^2-4n+6n-12+11}{n-2}=\dfrac{\left(n-2\right)\left(n^2+2n+6\right)+11}{n-2}=n^2+2n+6+\dfrac{11}{n-2}\)

là số nguyên => \(\dfrac{11}{n-2}\) là số nguyên => \(n-2\in\left\{-11,-1,1,11\right\}\) => \(n\in\left\{-9,1,3,13\right\}\) (thỏa mãn)

c)Vì \(\dfrac{n^2+2n+1}{n-3}=\dfrac{n^2-3n+5n-15+16}{n-3}=\dfrac{\left(n-3\right)\left(n+5\right)+16}{n-3}=n+5+\dfrac{16}{n-3}\)

là số nguyên => \(\dfrac{16}{n-3}\) là số nguyên => \(n-3\in\left\{-16,-8,-4,-2,-1,1,2,4,8,16\right\}\) => \(n\in\left\{-13,-5,-1,1,2,4,5,7,11,19\right\}\)

Vậy...

a)\(A=5x^2-7x-4=5\left(x^2-\dfrac{7}{5}x-\dfrac{4}{5}\right)=5\left(x^2-\dfrac{7}{5}x+\dfrac{49}{100}-\dfrac{129}{100}\right)=5\left[\left(x-\dfrac{7}{10}\right)^2-\dfrac{129}{100}\right]\ge5.-\dfrac{129}{100}=-\dfrac{129}{20}\)Dấu = xảy ra <=> \(x=\dfrac{7}{10}\)

b)\(B=6x^2+5x+4=6\left(x^2+\dfrac{5}{6}x+\dfrac{2}{3}\right)=6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}+\dfrac{71}{144}\right)=6\left[\left(x+\dfrac{5}{12}\right)^2+\dfrac{71}{144}\right]\ge\dfrac{71}{24}\)

Dấu = xảy ra <=> \(x=-\dfrac{5}{12}\)

c)\(C=2x^2+2x+2=2\left(x^2+x+1\right)=2\left(x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\right)=2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\ge\dfrac{3}{2}\)

Dấu = xảy ra <=> \(x=-\dfrac{1}{2}\)

d)\(D=8x^2-8x+7=8\left(x^2-x+\dfrac{7}{8}\right)=8\left(x^2-x+\dfrac{1}{4}+\dfrac{5}{8}\right)=8\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{8}\right]\ge5\)

Dấu = xảy ra <=> \(x=\dfrac{1}{2}\)

Ta có \(16x^4-y^4=9y^2+16\Rightarrow16x^4=y^4+9y^2+16\)

Khi đó, ta có \(y^4+9y^2+16\) là số chính phương. (1)

Mặt khác, ta có \(\left(y^2+5\right)^2-y^4-9y^2-16=y^2+9>0\)

Lại có \(y^4+9y^2+16-\left(y^2+4\right)^2=y^2\ge0\)

Do đó, ta có \(\left(y^2+4\right)^2\le y^4+9y^2+16< \left(y^2+5\right)^2\)

Khi đó, ta có \(y^4+9y^2+16=\left(y^2+4\right)^2\) (do (1))

\(\Leftrightarrow y^4+9y^2+16=y^4+8y^2+16\Leftrightarrow y=0\), thay vào, ta có x ∈ {1.-1}

Vậy...