x⁴ - 3x² +1

= x⁴ - 2x² + 1 - x²

= ( x² - 1 )² - x²

= ( x²- 1 +x ) ( x² - 1 - x ) 

umk, mình biết rồi. Tại đọc nhầm đề ấy mà

Bài 1.2

\(A=\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\)

C1:Bạn dùng pp chặn như bài 2.2

C2: (Gợi ý)\(\sqrt{x}+2\ge2\) và \(\sqrt{x}+2\inƯ\left(3\right)\)\(\Rightarrow\sqrt{x}+2=3\Leftrightarrow x=1\)

Vậy x=1 thì A nguyên

Bài 2.2

\(A=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\)

Do \(\sqrt{x}\ge0;\forall x\)\(\Rightarrow\sqrt{x}+2\ge2\) \(\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\)\(\Rightarrow A\le\dfrac{7}{2}\) (1)

mà \(\dfrac{5}{\sqrt{x}+2}>0;\forall x\Rightarrow A>1\) (2)

Từ (1) (2) \(\Rightarrow1< A\le\dfrac{7}{2}\) mà A nguyên

\(\Rightarrow\left[{}\begin{matrix}A=2\\A=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}1+\dfrac{5}{\sqrt{x}+2}=2\\1+\dfrac{5}{\sqrt{x}+2}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=5\\\sqrt{x}+2=\dfrac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

Bài 3.2

\(A=\dfrac{-x-2\sqrt{x}-5}{\sqrt{x}+2}\)\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=-\sqrt{x}-\dfrac{5}{\sqrt{x}+2}\)

\(=2-\left(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\right)\)

Áp dụng bđt cosi: \(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right).\dfrac{5}{\sqrt{x}+2}}=2\sqrt{5}\)

\(\Rightarrow A\le2-2\sqrt{5}\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}+2=\dfrac{5}{\sqrt{x}+2}\Leftrightarrow x=9-4\sqrt{5}\)

A = \(\dfrac{4\sqrt{x}+9}{2\sqrt{x}+1}\)

Mà \(4\sqrt{x}+9>0\)

\(2\sqrt{x}+1>0\)

=> A > 0

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}\) = \(2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 0 < A \(\le9\)

Mà A thuộc Z

<=> A \(\in\){1;2;3;4;5;6;7;8;9}

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}=2+\dfrac{7}{2\sqrt{x}+1}\)

Mà \(2\sqrt{x}+1>0< =>\dfrac{7}{2\sqrt{x}+1}>0\)

<=> A > 2

Có \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)

<=> \(A\le9\)

<=> 2 < A \(\le9\)

Mà A thuộc Z

<=> \(A\in\left\{3;4;5;6;7;8;9\right\}\)

Đến đây bn thay A vào để tìm x nhé

A = \(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}=2+\dfrac{4}{3\sqrt{x}+2}\)

Có \(3\sqrt{x}+2>0< =>\dfrac{4}{3\sqrt{x}+2}>0\) <=> A > 2

Có: \(3\sqrt{x}+2\ge2< =>\dfrac{4}{3\sqrt{x}+2}\le2\) <=> A \(\le4\)

<=> 2 < A \(\le4\)

Mà A nguyên

<=> \(\left[{}\begin{matrix}A=3\\A=4\end{matrix}\right.\)

TH1: A = 3

<=> \(\dfrac{4}{3\sqrt{x}+2}=1\)

<=> \(3\sqrt{x}+2=4< =>x=\dfrac{4}{9}\)

TH2: A = 4

<=> \(\dfrac{4}{3\sqrt{x}+2}=2< =>3\sqrt{x}+2=2< =>x=0\)

a=\(\sqrt{X^2+6+X}\)

pt\(\Leftrightarrow\)a²+6x-6=(2x+1)a

.....

(a-3)(a+2-2x)=0

...

giải 2 theo 2 trường hợp

♥♥♥

\(x^4-3x^2+1\)

\(=\left(x^4-2x^2+1\right)-x^2\)

\(=\left(x^2-1\right)^2-x^2=\left(x^2-1-x\right)\left(x^2-1+x\right)\)

x4 - 3x2 +1

= (x2)2 - 3x2 + x2 + 1

= (x2 + 1)2 + x2

= (x2 + 1 + x). (x2 + 1 - x)

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=u\\\frac{1}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{3}{5}u+v=10\\\frac{3}{4}u+\frac{3}{4}v=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3u+5v=50\\3u+3v=48\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}v=1\\u=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=15\\\frac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{15}\\y=1\end{matrix}\right.\)