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Cho biểu thức
A= (\( {1 \over x-2}\)+\({1 \over x+2}\)) : \( {5-x \over x-2}\)
a) Tìm ĐKXĐ
b) Rút gọn A
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
\(1.\)
\(a.=3\left(x+2\right)\)
\(b.=4\left(x-y\right)+x\left(x-y\right)\)
\(=\left(4+x\right)\left(x-y\right)\)
\(c.=\left(x-6\right)\left(x+6\right)\)
\(d.=\left(x^2-2y^2\right)\left(x^2+2y^2\right)\)
\(2.\)
\(a.ĐKXĐ:\)\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
\(b.A=\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3}{x+1}với\)\(x\ne\pm1\)
\(c.A=-1\Leftrightarrow\frac{3}{x+1}=-1\)
\(\Rightarrow\left(x+1\right).-1=3\)
\(-x-1=3\)
\(-x=4\)
\(\Rightarrow x=4\left(t/mđk\right)\)
\(d.\)Để \(x\in Z,A\in Z\Leftrightarrow x+1\inƯ\left(3\right)\)
\(Ư\left(3\right)\in\left\{\pm1,\pm3\right\}\)
| x+1 | 1 | -1 | 3 | -3 |
| x | 0 | -2 | 2 | -4 |
Vậy \(x\in\left\{0,-2,2,-4\right\}\)
1a) 3x + 6 = 3 (x + 2)
b) 4x - 4y + x2 - xy = (4x - 4y) + (x2 - xy) = 4 (x - y) + x (x - y) = (4 + x) (x - y)
c) x2 - 36 = x2 - 62 = (x + 6) (x - 6)
2a) phân thức A được xác định khi \(x^2-1\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\ne0\)
\(\Rightarrow x+1\ne0..và..x-1\ne0\)
\(x\ne-1..và..x\ne1\)
b) \(A=\frac{3x-3}{x^2-1}=\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{3}{x+1}\)
c) \(A=-1\Rightarrow\frac{3}{x+1}=-1\)
\(\Rightarrow x+1=-3\)
\(x=-4\left(TM\text{Đ}K\right)\)
Vậy x = -1 thì A = -1
#Học tốt!!!
~NTTH~
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
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