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A= (4x2 + y2).[(2x)2 - y2] = (4x2 +y2)(4x2 - y2) = (4x2)2 _ (y2)2 = 16x4 - y4
\(1.\)
\(a.=3\left(x+2\right)\)
\(b.=4\left(x-y\right)+x\left(x-y\right)\)
\(=\left(4+x\right)\left(x-y\right)\)
\(c.=\left(x-6\right)\left(x+6\right)\)
\(d.=\left(x^2-2y^2\right)\left(x^2+2y^2\right)\)
\(2.\)
\(a.ĐKXĐ:\)\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
\(b.A=\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3}{x+1}với\)\(x\ne\pm1\)
\(c.A=-1\Leftrightarrow\frac{3}{x+1}=-1\)
\(\Rightarrow\left(x+1\right).-1=3\)
\(-x-1=3\)
\(-x=4\)
\(\Rightarrow x=4\left(t/mđk\right)\)
\(d.\)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~
- x2.(x3-x2+x-1)
- x.( x3-3x2-1)+3
- x.(x2-xy-y2)
Tìm x:
x3-16x = 0
=> x.(x2-16) = 0
=> x = 0 hay x2-16 = 0
=> x = 0 hay x2 = 0+16
=> x = 0 hay x2 = 16
=> x = 0 hay x = 4 hay x = -4
Bài làm:
a) đkxđ: \(x\ne y\ne0\)
Ta có: \(\frac{x^2-xy}{3x^2-3xy}=\frac{x\left(x-y\right)}{3x\left(x-y\right)}=\frac{1}{3}\)
b) đkxđ: \(x\ne\pm4\)
Ta có: \(\left(\frac{1}{x+4}+\frac{8}{x^2-16}\right).\frac{x+1}{x-4}\)
\(=\frac{x-4+8}{\left(x+4\right)\left(x-4\right)}.\frac{x+1}{x-4}\)
\(=\frac{x+4}{\left(x+4\right)\left(x-4\right)}.\frac{x+1}{x-4}=\frac{x+1}{\left(x-4\right)^2}\)
a) ĐKXĐ : \(x\ne y\ne0\)
\(\frac{x^2-xy}{3x^2-3xy}=\frac{x\left(x-y\right)}{3x\left(x-y\right)}=\frac{x}{3x}=\frac{1}{3}\)
b) ĐKXĐ : \(x\ne\pm4\)
\(\left(\frac{1}{x+4}+\frac{8}{x^2-16}\right)\cdot\frac{x+1}{x-4}\)
\(=\left(\frac{1}{x+4}+\frac{8}{\left(x+4\right)\left(x-4\right)}\right)\cdot\frac{x+1}{x-4}\)
\(=\left(\frac{x-4}{\left(x+4\right)\left(x-4\right)}+\frac{8}{\left(x+4\right)\left(x-4\right)}\right)\cdot\frac{x+1}{x-4}\)
\(=\frac{x+4}{\left(x+4\right)\left(x-4\right)}\cdot\frac{x+1}{x-4}\)
\(=\frac{x+1}{\left(x-4\right)^2}\)