Phép nhân và phép chia các đa thức
Các câu hỏi tương tự
Rút gọn các biểu thức sau:
A= \(\left(x+1\right).\left(x^2-x+1\right)+2.\left(x+1\right)-x.\left(x^2+2\right).\)
B= \(\left(5x+1\right).\left(x+7\right)-5x.\left(x-1\right).\)
1. frac{1}{2}x^2-left(frac{1}{2}x-4right)frac{1}{2}x-14
2. 3left(1-4xright)left(x-1right)+4left(3x-2right)left(x+3right)-27
3. 6xleft(5x+3right)+3xleft(1-10xright)7
4. left(3x-3right)left(5-21xright)+left(7x+4right)left(9x-5right)44
5. left(-2+x^3right)left(-2+x^2right)left(-2+x^2right)1
Đọc tiếp
1. \(\frac{1}{2}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
2. \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
3. \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
4. \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
5. \(\left(-2+x^3\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
PHƯƠNG PHÁP THÊM BỚT HẠNH TỬ VÀ PHƯƠNG PHÁP ĐỔI BIẾN :
a) 4x^4+1
b) x^8+4
c) x^4+x^2+1
d) x^7+x^5+1
e) x^7+x^5-1
f) left(x+2right)left(x+4right)left(x+6right)left(x+8right)+16
g) left(x+2right)left(x+3right)left(x+4right)left(x+5right)-24
h) left(x-1right)left(x-3right)left(x-5right)left(x-7right)-20
i) left(x-1right)left(x+2right)left(x+3right)left(x+6right)-28
Đọc tiếp
PHƯƠNG PHÁP THÊM BỚT HẠNH TỬ VÀ PHƯƠNG PHÁP ĐỔI BIẾN :
a) \(4x^4+1\)
b) \(x^8+4\)
c) \(x^4+x^2+1\)
d) \(x^7+x^5+1\)
e) \(x^7+x^5-1\)
f) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
g) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
h) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
i) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)
Rút gọn biểu thức sau:
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^2+3\left(x-1\right)\left(x+1\right)\)
b, \(\left(x^4-5x^2+25\right)\left(x^2+5\right)-\left(2+x^2\right)^2+3\left(1+x^2\right)^2\)
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
Tìm x biết :a) left(x-2right)^3+6left(x+1right)^2-x^3+120b) left(x-5right)left(x+5right)-left(x+3right)^3+3left(x-2right)^2left(x+1right)^2-left(x+4right)left(x-4right)+3x^2c) left(2x+3right)^2+left(x-1right)left(x+1right)5left(x+2right)^2-left(x-5right)left(x+1right)+left(x+4right)^2d) left(1-3xright)^2-left(x-2right)left(9x+1right)left(3x-4right)left(3x+4right)-9left(x+3right)^2
Đọc tiếp
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
Rút gọn các biểu thức :
a, \(\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)\)
b, \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(c,\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
Rút gọn Biểu thức :
a) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
b) \(3\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
3) frac{x-2}{x-5}-frac{5}{x^2-5x}frac{1}{x}
Leftrightarrowfrac{x-2}{x-5}-frac{5}{x.left(x-5right)}frac{1}{x}
Leftrightarrowfrac{x.left(x-2right)}{x.left(x-5right)}-frac{5}{x.left(x-5right)}frac{1.left(x-5right)}{x.left(x-5right)}
Mc: x.left(x-5right)
Leftrightarrow x^2 - 2x - 5 x - 5
Leftrightarrow x^2 - 2x - x - 5 + 5 0
Leftrightarrow x^2 - 3x 0
Leftrightarrow x . (x - 3) 0
Leftrightarrow x 0 hoặc x - 3 0
Leftrightarrow x 0 hoặc x 3
Vậy x 0 hoặc x 3
x-5ne0Rightarrow xn...
Đọc tiếp
3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)
Mc: \(x.\left(x-5\right)\)
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0
\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0
\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3
Vậy \(x\) = 0 hoặc \(x\) = 3
\(x-5\ne0\Rightarrow x\ne5\)
\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {3}
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
Mc: \(x.\left(x+7\right)\)
\(\Leftrightarrow x^2-4x-x-7=-7\)
\(\Leftrightarrow x^2-4x-x=-7+7\)
\(\Leftrightarrow\) \(x^2-5x=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x-5=0\)
\(\Leftrightarrow x=0\) hoặc \(x=5\)
Vậy \(x=0\) hoặc \(x=5\)
\(x+7\ne0\Rightarrow x\ne-7\)
\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {5}
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Mc : \(\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)
\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)
\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=2\) hoặc \(x=2\)
\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
MC: \(\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)
\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4.\left(x-1\right)=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)
\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(Mc:\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)
\(\Leftrightarrow0=0\) (Nhận)
Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)