bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

\(A=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

a) ĐKXĐ : x ≠ ±2

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

b) Để A = 1 => \(\frac{1}{x-2}=1\)=> x - 2 = 1 => x = 3 ( tm )

c) Để A > 1 => \(\frac{1}{x-2}>1\)

=> \(\frac{1}{x-2}-1>0\)

=> \(\frac{1}{x-2}-\frac{x-2}{x-2}>0\)

=> \(\frac{1-x+2}{x-2}>0\)

=> \(\frac{-x+3}{x-2}>0\)

Xét hai trường hợp

1. \(\hept{\begin{cases}-x+3>0\\x-2>0\end{cases}}\Rightarrow\hept{\begin{cases}-x>-3\\x>2\end{cases}}\Rightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}}\Rightarrow2< x< 3\)

2. \(\hept{\begin{cases}-x+3< 0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}-x< -3\\x< 2\end{cases}}\Rightarrow\hept{\begin{cases}x>3\\x< 2\end{cases}}\)( loại )

Vậy với 2 < x < 3 thì A > 1

d) Để A nguyên => \(\frac{1}{x-2}\)nguyên

=> 1 ⋮ x - 2

=> x - 2 ∈ Ư(1) = { ±1 }

=> x ∈ { 1 ; 3 } thì A nguyên

a) \(ĐKXĐ:x\ne\pm2\)

\(A=\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{6-5x}{x^2-4}\)

\(\Leftrightarrow A=\dfrac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

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

\(\Leftrightarrow A=\dfrac{1}{x-2}\)

b) Để A = 1

\(\Leftrightarrow\dfrac{1}{x-2}=1\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow x=3\) (tm)

Vậy ...

c) Để A > 1

\(\Leftrightarrow\dfrac{1}{x-2}>1\)

\(\Leftrightarrow\dfrac{1}{x-2}-1>0\)

\(\Leftrightarrow\dfrac{1-x+2}{x-2}>0\)

\(\Leftrightarrow\dfrac{-x+3}{x-2}>0\)

\(\Leftrightarrow\left(3-x\right)\left(x-2\right)>0\)

Trường hợp \(\left\{{}\begin{matrix}3-x>0\\x-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x>2\end{matrix}\right.\)

\(\Leftrightarrow2< x< 3\) (tm)

Trường hợp \(\left\{{}\begin{matrix}3-x< 0\\x-2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>3\\x< 2\end{matrix}\right.\) (ktm)

Vậy ...

d) Để A nguyên 

\(\Leftrightarrow\dfrac{1}{x-2}\in Z\)

\(\Leftrightarrow x-2\inƯ\left(1\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{1;3;0;4\right\}\)

Vậy ...

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

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

viết lại biểu thức 

a) \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}=\dfrac{\left(x-1-2\right)\left(x-1+2\right)}{\left(2x+1-x-2\right)\left(2x+1+x+2\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{3\left(x-1\right)\left(x+1\right)}\)  (1)

\(\Rightarrow\) ĐKXĐ: \(x\ne\pm1\)

b) \(\left(1\right)=\dfrac{x-3}{3x-3}\) (2)

c) Thay \(x=-3;x=1\) vào (2) ta có: \(\left\{{}\begin{matrix}B=\dfrac{-3-3}{3.\left(-3\right)-3}=\dfrac{1}{2}\\B=\dfrac{1-3}{3.1-3}=0\end{matrix}\right.\)

d) \(B=5\Rightarrow\dfrac{x-3}{3x-3}=5\Leftrightarrow x-3=15x-15\Leftrightarrow x=\dfrac{6}{7}\)

a) a ≠ 0 ,    a ≠   − 5  

b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2  

c) Thay a = -1 (TMĐK) vào a ta được A = -1

d) Ta có A = 0 Û a = 1 (TMĐK)

a) x ≠ 0 ,    x ≠     − 2  

b) Ta có D = x 2  - 2x - 2.

c) Chú ý D = - x 2 - 2x - 2 = - ( x   +   1 ) 2  - 1 ≤ -1. Từ đó tìm được giá trị lớn nhất của D = -1 khi x = -1.

để A xác định

\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)

\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)

\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)

\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)

C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)

d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)

\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)

\(x^2-4\inℕ\)

\(\Rightarrow x^2\in\left\{0;4;12\right\}\)

Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn

mk học lớp 6 mong mn thông cảm nếu có sai sót

Mk nhầm nha

=2/x-2 nha bạn

b, =1/-3

c, 2 chia hết cho x-1

=> x-1 E {-2;-1;1;2}

=> x E {-1;0;2;3} mk có tính nhầm 1 chỗ