BÀI 5 : CHO
E=\(\frac{x^2+6x+9}{x^3+3x^2-27x+27}\). \(\left(\frac{x^2+6x+9}{x^3+3x^2-27x+27}+\frac{2}{3x}:\left(\frac{1}{x}+\frac{1}{3}\right)^2\right)\)
F=\(\frac{3+x}{3-x}\) .\(\frac{x^2-6x+9}{9x^2}\).\(\left(\frac{3}{3-x}-\frac{9}{27+x^3}.\frac{x^2-3x+9}{3-x}\right)\)
a, RÚT GỌN E VÀ F
Giải các bất phương trình
a)
\(\left(\frac{x+1}{3x^2+3x}+\frac{1-2x}{6x^2-3x}-1\right):\frac{1-x}{2x}\le1\)
b) \(x^3-9x^2+27x-19\le0\)
c) \(\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+7\right)>297\)
d) \(\left(x+2\right)^3>1+x+x^2+x^3\)
Mn giúp mk với ạ. Mk cần gấp lắm. Xie xie, cảm ơn nhiều ạ(cuối đầu)
Phân tích các đa thức sau thành nhân tử:
\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)
\(10x-25-x^2\)
\(\frac{1}{25}x^2-64y^2\)
\(x^3+\frac{1}{27}\)
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(8x^3+12x^2y+6xy+y^3\)
\(-x^3+9x^2-27x+27\)
Tìm x, biết:
\(4\left(x^2+x+1\right)^3=27x^2\left(x-1\right)^2+\left(x-1\right)^2\left(2+x\right)\left(2x+1\right)^2\)
Bài 1 : dùng hẳng đẳng thức để khai triển và thu gọn
a) \(\left(2x^2+\frac{1}{3}\right)^3\)
b) \(\left(2x^2y-3xy\right)^3\)
c) \(\left(-3xy^4+\frac{1}{2}x^2y^2\right)^3\)
d) \(\left(-\frac{1}{3}ab^2-2a^3b\right)^3\)
e) \(\left(x+1\right)^3-\left(x-1\right)^3-6.\left(x-1\right).\left(x+1\right)\)
f) \(x.\left(x-1\right).\left(x+1\right)-\left(x+1\right).\left(x^2-x+1\right)\)
g) \(\left(x-1\right)^3-\left(x+2\right).\left(x^2-2x+4\right)+3.\left(x-4\right).\left(x+4\right)\)
h) \(3x^2.\left(x+1\right).\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right).\left(x^4+x^2+1\right)\)
k) \(\left(x^4-3x^2+9\right).\left(x^2+3\right)+\left(3-x^2\right)^3-9x^2.\left(x^2-3\right)\)
l) \(\left(4x+6y\right).\left(4x^2-6xy+9y^2\right)-54y^3\)
bt em gửi cô Thương
1)\(ĐKXĐ\hept{\begin{cases}x\ne1\\x\ne3\end{cases}}\)
\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)
\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-x-3x+3}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{x\left(x-1\right)-3\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x-6}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)( tm)
Vậy nghiemj của pt x=3
2)\(x^3-x^2-9x+9=0\)
\(\Leftrightarrow x^2\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)hoặc x+3=0
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)hoặc x=-3
Vậy tập hợp nghiệm \(S=\left\{1;3;-3\right\}\)
Chứng minh đẳng thức :
a)\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)=\frac{3}{3-x}\)
b) \([\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)]:\frac{x-1}{x}=\frac{2x}{x-1}\)
Rút gọn biểu thức sau: A=\(\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right].\frac{4x^2+4x+1}{\left(x+4\right)\left(3-x\right)}\)
Phân tích các đa thức sau thành nhân tử
\(2x^3-x^2+5x+3\)
\(27x^3-27x^2+18x-4\)
\(x^2+2xy+y^2-x-y-12\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(x^3+3xy+y^3-1\)
\(4x^4+4x^3+5x^2+2x+1\)
\(x^8+x+1\)
\(x^8+3x^4-4\)