em mới lớp 6 nên chưa biết

\(B=\sqrt{x-4\sqrt{x}+4}+\sqrt{x-6\sqrt{x}+9}\)

    \(=\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}\)

    \(=\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)

Dấu "=" <=> 4 < x < 9

\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\\ =\dfrac{x+2-\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}+1-x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x+2-2x+4\sqrt{x}-\sqrt{x}-1+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)^2}\)

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

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

\(A=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)

\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)

\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}+2}\)

\(=\dfrac{5\left(4\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

\(\dfrac{6}{x}+8y\) hay \(\dfrac{6}{x}+\dfrac{8}{y}\)? Nếu là 8y tại sao ko cộng luôn với 2y thành 10y nhỉ?

đề ra thế 

cái tứ 1

b: Để A là số nguyên thì \(2x+2⋮x+3\)

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

hay \(x\in\left\{-4;-1;-5;1;-7\right\}\)

\(P=\dfrac{6}{x}+\dfrac{3}{2}x+\dfrac{24}{y}+\dfrac{3}{2}y-\dfrac{1}{2}\left(x+y\right)\ge2\sqrt{6.\dfrac{3}{2}}+2\sqrt{24.\dfrac{3}{2}}-\dfrac{1}{2}.6=15\Rightarrow min=15\Leftrightarrow x=2;y=4\)

a: ĐKXĐ: \(x\notin\left\{4\right\}\)

x²-3x=0

=>x(x-3)=0

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

Thay x=0 vào A, ta được:

\(A=\dfrac{0-5}{0-4}=\dfrac{-5}{-4}=\dfrac{5}{4}\)

Thay x=3 vào A, ta được:

\(A=\dfrac{3-5}{3-4}=\dfrac{-2}{-1}=\dfrac{2}{1}=2\)

b: \(B=\dfrac{x+5}{2x}-\dfrac{x-6}{5-x}-\dfrac{2x^2-2x-50}{2x^2-10x}\)

\(=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x\left(x-5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x-5\right)}=\dfrac{x-5}{2x}\)

c: Đặt P=A:B

ĐKXĐ: \(x\notin\left\{4;5;0\right\}\)

P=A:B

\(=\dfrac{x-5}{x-4}:\dfrac{x-5}{2x}\)

\(=\dfrac{x-5}{x-4}\cdot\dfrac{2x}{x-5}=\dfrac{2x}{x-4}\)

Để P là số nguyên thì \(2x⋮x-4\)

=>\(2x-8+8⋮x-4\)

=>\(8⋮x-4\)

=>\(x-4\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(x\in\left\{5;3;6;2;8;0;12;-4\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;6;2;8;12;-4\right\}\)

Bài 3: Cho biểu thức A = x - 5/x - 4 và B = x + 5/2x - x - 6/5 - x - 2x² - 2x - 50 / 2 x^2 - 10x t

Ta có x² - 3x = 0 suy ra x x (x - 3) = 0

x = 0; x = 3

Với x = 0 suy ra A = 5/4 v

Với x = 3 suy ra A = 2

Để p đạt giá trị nguyên khi 8/x - 4 cũng phải có giá trị nguyên 28 : (x - 4)

Vậy x - 4 thuộc ước chung của 8 = -8, -4, -1, 1, 4, 8

x - 4 = 8 suy ra x = 4

x - 4 = 4 suy ra 2x = 0 loại

x - 4 = -1 suy ra x = 3 thỏa mãn

x - 4 = 1 suy ra x = 5 loại

x - 4 = 4 - 2x = 8 thỏa mãn

x - 4 = 8 suy ra x = 12 thỏa mãn