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\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)
\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)
Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)
c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z
=> -2x ⋮ x + 1
=> -2x - 2 + 2 ⋮ x + 1
=> -2( x + 1 ) + 2 ⋮ x + 1
Vì -2( x + 1 ) ⋮ ( x + 1 )
=> 2 ⋮ x + 1
=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }
a) \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(A=\left(\frac{x}{\left(x-2\right)\cdot\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(A=\frac{x+x-2-2\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{x+2-x}{x+2}\)
\(A=\frac{2x-2-2x-4}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{2}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\frac{x+2}{2}\)
\(\Rightarrow A=\frac{-3}{x-2}\)
b) Với x = -4 . Ta có :
\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{-3}{-6}=\frac{1}{2}\)
cho tam giác ABC có 3 góc nhọn , 2 đường cao BE và CF cắt nhau tại H
a/ Chứng minh tam giác AEB ~ tam giác AFC
b/ chứng minh tam giác DEF ~ tam giác ABC
c/ Tia AH cắt BC tại D. Chứng minh FC là tia phân giác góc DFE ?
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
a) Điều kiện : \(x\ne2;x\ne3\)
\(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Điều kiện \(x\in Z;x\ne2;x\ne3\)
Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên
\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)
mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
a ) \(A=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)
\(=\frac{x+2-\left(x-2\right)+x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+8}{x^2-4}\)
b ) \(A=\frac{x^2+8}{x^2-4}=\frac{\left(x^2-4\right)+12}{x^2-4}=1+\frac{12}{x^2-4}\)
Để \(A\in Z\Leftrightarrow12⋮x^2-4\)
\(x^2-4\inƯ\left(12\right)=\left\{-12;-6;-4;-2;-1;1;2;4;6;12\right\}\)
Xét từng thường hợp của x ta tìm đc : \(x=\left\{-4;0;4\right\}\)
\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)
= \(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+2^2}{x^2-2^2}\)
= \(\frac{4}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+2^2}{x^2-2^2}\)
=\(\frac{4}{x^2-2^2}+\frac{x^2+2^2}{x^2-2^2}\)
= \(\frac{4+x^2+2^2}{x^2-2^2}\)