Bài 1:
a, \(\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)
b, \(\frac{2}{x^2+2xy}và\frac{1}{xy+2y^2}\)
c, \(\frac{5}{6-2x}và\frac{3}{x^2-9}\)
\(\frac{x}{4x^2y^3}\)và \(\frac{5}{6x^3y}\)
\(\frac{5}{6-2x}\)và\(\frac{3}{x^2-9}\)
\(\frac{2}{x^2+2xy}\)và\(\frac{1}{xy+2y^2}\)
các anh chị giúp em nha mài em phải làm bài rồi
Giải hpt:
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+x^2y+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^4+2x^2y+x^2y^2=-2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
3) ta xét phương trình thứ nhất
\(x-\frac{1}{x}=y-\frac{1}{y}\)
<=>\(x-y-\frac{1}{x}+\frac{1}{y}=0\)
<=>\(x-y-\left(\frac{1}{x}-\frac{1}{y}\right)=0\)
<=>\(x-y-\left(\frac{y-x}{xy}\right)=0\)
<=>\(\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\)
<=>\(x=y\) hoặc xy=-1
Với x=y thay vào phương trình thứ hai ta có
\(2x=x^3+1
\)
<=> \(x^3-2x+1=0\)
<=>\(x^3-x^2+x^2-x-x+1=0\)
<=>\(\left(x-1\right)\left(x^2+x-1\right)=0\)
<=> \(x=1\) hoặc \(x^2+x-1=0\)
\(x^2+x-1=0\) <=> \(x=\frac{-1+\sqrt{5}}{2}\)
hoặc \(x=\frac{-1-\sqrt{5}}{2}\)
Đối với xy=-1 thì y=-1/x thay vào phương trình 2 giải bình thường
bài 1 : thu gọn đa thức , tìm bậc , hệ số cao nhất
A = 15x^2y^3 + 7x^2 - 8x^3y^2 - 12x^2 + 11x^3y^2 - 12x^2y^3
B = 3x^5y + \(\frac{1}{3}\)xy^4 + \(\frac{3}{4}\)x^2y^3 - \(\frac{1}{2}\)x^5y + 2xy^4 - x^2y^3
bài 2 : tính giá trị biểu thức
A = 3x^3y + 6x^2y^2 + 3xy^3 tại x = \(\frac{1}{2}\); y = -\(\frac{1}{3}\)
B = x^2y^2 + xy +x^3 + y^3 tại x = -1 ; y = 3
bài 3 : cho đa thức
P(x) = x^4 + 2x^2 + 1
Q(x) = x^4 + 4x^3 + 2x^2- 4x + 1
tính P(-1); P(\(\frac{1}{2}\)) ; q(-2);Q(1)
bài 4 : tìm hệ số a của đa thức M(x)= ax^2 + 5x - 3 , tại M (-3) = 0
bài 5 : tìm các hệ số a , b của đa thức f(x) = ax + b , biết f(2) = 3 ; f(-1) = 9
Tìm đa thức M , biết :
a) \(M-\left(\frac{1}{2}x^2y-5xy^2+x^3-y^3\right)=\frac{3}{4}xy^2-2x^2y+\)\(2y^3-\frac{1}{3}x^3\)
b)\(\left(-\frac{1}{3}x^3y^3+5x^2y^2-\frac{5}{2}xy\right)-M=xy-\frac{1}{6}x^3y^3-3x^2y^2\)
c)\(\left(\frac{2}{7}xy^4-5x^5+7x^2y^3-3\right)+M=0\)
Tìm x , y , z :
a) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và 2x + 3y - z = 50
b) \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{x-5}{6}\)và 5x - 3y - 4z = 46
c) \(\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\)và x + y + z = 107
d) \(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\frac{4y-5z}{6}\)và 3x - 2y + 5z = 96
a
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)
Thay vào,ta được:
\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow4k+2+9k+6-4k-3=50\)
\(\Leftrightarrow9k+5=50\)
\(\Leftrightarrow9k=45\)
\(\Leftrightarrow k=5\)
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)
\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)
\(\Rightarrow x=2\cdot2+1=5\)
\(y=4\cdot2-3=5\)
\(z=2\cdot6+5=17\)
Câu c tương tự như câu 1
\(c,\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\)và x + y + z = 107
Ta có : \(\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\Leftrightarrow\frac{x}{\frac{5}{2}}=\frac{y}{\frac{10}{3}}=\frac{z}{12}=\frac{x+y+z}{\frac{5}{2}+\frac{10}{3}+12}=\frac{107}{\frac{107}{6}}=107\cdot\frac{6}{107}=6\)
Vậy : \(\hept{\begin{cases}\frac{2x}{5}=6\\\frac{3y}{10}=6\\\frac{z}{12}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=15\\x=20\\z=72\end{cases}}\)
8,Thực hiện phép tính
a,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
b,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
d,\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
e,\(\frac{2x+y}{2x^2-xy}+\frac{16x}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
f,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
Tính
a) \(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)\)
b) \(\frac{x^3-3x^2+2x}{3x^2-4x+1}.\left(\frac{x-1}{x}-\frac{2x-6}{x-1}+\frac{x+1}{x-2}\right)\)
c) \(\frac{3x-3y}{2x^2-2xy+2y^2}:\frac{6x^2-12xy+6y^2}{5x^3+5y^3}:\frac{5x}{x-y}\)
a)\(ĐKXĐ:x\ne0;-1\)
Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)
Tính:
\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\)
\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(\)Thank you so much!
a.\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\) MTC=6x2y
\(=\frac{3\left(4x-1\right)}{6x^2y}-\frac{2\left(7x-1\right)}{6x^2y}\)
\(=\frac{12x-3-\left(14x-2\right)}{6x^2y}\)
\(=\frac{12x-3-14x+2}{6x^2y}\)
\(=\frac{-2x-1}{6x^2y}=\frac{2\left(-x-1\right)}{6x^2y}=-\frac{x-1}{3x^2y}\)
b.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) MTC= 2x (x + 3)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c.\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)MTC= xy (x+2y).(x-2y)
\(=\frac{2xy\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{xy\left(x+2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{3x^2y-2xy^2+4xy}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{xy\left(3x-2y+4\right)}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
Chọn mk nha!
Bài 1: rút gọn phân thức
a) \(\frac{14xy^2\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
b) \(\frac{8xy\left(3x-1\right)^2}{12x^3\left(1-3x\right)}\)
c) \(\frac{20x^2-45}{\left(2x+3\right)^2}\)
d) \(\frac{5x^2-10xy}{2\left(2y-x\right)^3}\)
e) \(\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
f) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}\)
g) \(\frac{32x-8x^2+2x^3}{x^3+64}\)
h) \(\frac{5x^3+5x}{x^4-1}\)
Bài 2: Quy đồng mẫu thức của các phân thức sau
a) \(\frac{7x-1}{2x^2+6x};\frac{5-3x}{x^2-9}\)
b) \(\frac{x+1}{x-x^2};\frac{x+2}{2-4x+2x^2}\)
c) \(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1};\frac{6}{x-1}\)
d) \(\frac{7}{5x};\frac{4}{x-2y};\frac{x-y}{8y^2-2x^2}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)