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a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
a) P có nghĩa khi \(\hept{\begin{matrix}2x+4\ne0\\2x-4\ne0\\x^2-4\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow\hept{\begin{matrix}2\left(x+2\right)\ne0\\2\left(x-2\right)\ne0\\\left(x-2\right)\left(x+2\right)\ne0\\x-2\ne0\end{matrix}\Leftrightarrow\hept{\begin{matrix}x+2\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow x\ne\pm2}\)
vậy P có nghĩa khi \(x\ne\pm2\)
b) \(P=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}-\frac{8}{x^2-4}\right):\frac{4}{x-2}\left(x\ne\pm2\right)\)
\(\Leftrightarrow P=\left(\frac{x+2}{2\left(x-2\right)}+\frac{x-2}{2\left(x+2\right)}-\frac{8}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\left[\frac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\left[\frac{x^2+4x+4}{2\left(x-2\right)\left(x+2\right)}+\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}\)
\(\Leftrightarrow P=\frac{2x^2-8}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}=\frac{2\left(x^2-4\right)\left(x-2\right)}{8\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(x-2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{x-2}{4}\)
vậy P=\(\frac{x-2}{4}\left(x\ne\pm2\right)\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)
Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên
=> 7 ⋮ x - 3
=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }
| x-3 | 1 | -1 | 7 | -7 |
| x | 4 | 2 | 10 | -4 |
So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn
Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên
(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.
( ác ) là từ ( các )
(gia strij) là từ ( giá trị )
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
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 \(x\in\left\{4;10;-4\right\}\)