Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

Bài 2:

\(x+y+1=0\Rightarrow x+y=-1\)

A = \(x\)(\(x+y\)) - y².(\(x+y\)) + \(x^2\) - y² + 2(\(x+y\)) + 3

Thay \(x\) + y  = -1 vào biểu thức A ta có:

A = \(x\).( -1) - y² .(-1) + \(x^2\)  - y² + 2(-1) + 3

A = -\(x\) + y² + \(x^2\) - y² - 2 + 3

A = \(x^2\) - \(x\) + 1

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

Để biểu thức nguyên thì \(3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\)

hay x=1

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

mà \(x>0=>\sqrt{x}+2>2\) nên \(\sqrt{x}+2=\left\{3\right\}=>x=1\left(tm\right)\)

Vaayy.....

Để biểu thức \(\dfrac{3}{\sqrt{x}+2}\) nguyên thì \(3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\)

hay x=1

`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`

   `A=4/[12-4]=1/2`

Với `x > 0,x ne 1` có:

`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`

`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`

`b)B=2/5A`

`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`

`<=>5\sqrt{x}-5=\sqrt{x}`

`<=>\sqrt{x}=5/4`

`<=>x=25/16` (t/m)

a) Ta có: \(P=\left(\dfrac{3}{x+1}+\dfrac{x-9}{x^2-1}+\dfrac{2}{1-x}\right):\dfrac{x-3}{x^2-1}\)

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

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

\(=\dfrac{2x-14}{x-3}\)

b) Ta có: \(x^2-9=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)

Thay x=-3 vào biểu thức \(P=\dfrac{2x-14}{x-3}\), ta được:

\(P=\dfrac{2\cdot\left(-3\right)-14}{-3-3}=\dfrac{-20}{-6}=\dfrac{10}{3}\)

Vậy: Khi \(x^2-9=0\) thì \(P=\dfrac{10}{3}\)

c) Để P nguyên thì \(2x-14⋮x-3\)

\(\Leftrightarrow2x-6-8⋮x-3\)

mà \(2x-6⋮x-3\)

nên \(-8⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(-8\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

\(\Leftrightarrow x\in\left\{4;2;5;1;7;-1;11;-5\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;7;11;-5\right\}\)

Vậy: Để P nguyên thì \(x\in\left\{4;2;5;7;11;-5\right\}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8

1) Xét rằng x > 7 <=> A < 0

Lại xét x < 7 thì mẫu là một số nguyên dương. P/s A có tử và mẫu đều là số dương, mà tử lại bất biến

A(max) <=> mẫu 7 - x nhỏ nhất <=> 7 - x = 1 => x = 7 - 1 = 6 <=> A = 1

Từ những điều trên thì A sẽ có GTLN khi và chỉ khi x = 6

\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)

\(A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Kết hợp đk

\(\Rightarrow x\in\left\{4;16;64\right\}\)