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QN
2
16 tháng 12 2018
ĐKXĐ : \(x\ne2\)
\(\frac{3x^2+6x+12}{x^3-8}\)
\(=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{3}{x-2}\)
3 tháng 12 2018
a) Để phân thức trên xác định \(\Leftrightarrow x^3-8\ne0\Leftrightarrow x\ne2\)
b) \(\frac{3x^2+6x+12}{x^3-8}\)
\(=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{3}{x-2}\)
NN
0
6 tháng 1 2019
câu 1:
x3-1+3x2-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)
6 tháng 1 2019
Câu 2 :
a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)
\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)
\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right):\left(x^2-1\right)\)
\(=x^2-2x+1\)
b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)
\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)
Câu 3 :
Sửa đề :
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
30 tháng 11 2019
1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0
Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)
b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)
B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x+3y}{x\left(x-3y\right)}\)
30 tháng 11 2019
\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)
\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)
8 tháng 5 2018
a,ĐKXĐ \(x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)
b,\(\Leftrightarrow3x^2+6x+12=0\)
\(\Leftrightarrow3\left(x^2+2x+1\right)+9=0\)
\(\Leftrightarrow3\left(x+1\right)^2+9=0\)(VÔ LÝ VÌ 3(x+1)2>=0 =>3(x+1)2+9>0)
vì vây ko có giá trị x để F =0
C, VỚI ĐKXĐ trên ,ta có
\(F=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{3}{x-2}\)
QA
21 tháng 7 2021
Trả lời:
a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)
b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)
c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)
d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)
e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
ra \(\frac{3}{x-2}\) đúng k ạ............