\(Q=\frac{2}{x}+\frac{3}{y}+\frac{6}{3x+2y}\)
cho \(xy=6;x>0;y>0\)
tìm gtnn của Q
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,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
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)\(\left(\frac{2}{7}xy^4-5x^5+7x^2y^3-3\right)+M=0\)
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!
1,Thực hiện phép tính
a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
b,\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
c,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
a) \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3x-2x+4x^2-2x-4x^2+4x-4x+4}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{2x\left(x-1\right)}\)
b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)
\(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)
c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)
\(=\frac{5x^2-y-3x^2+2xy}{xy}\)
\(=\frac{2x^2-y+2xy}{xy}\)
d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}\)
\(=\frac{2x+6}{2x\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{2}{2x}=\frac{1}{x}\)
Tính:
a) \(\frac{4x-1}{3x^2y}\)- \(\frac{7x-1}{3x^2y}\)
b) \(\frac{3}{2x+6}\)- \(\frac{x-6}{2x^2+6x}\)
c)\(\frac{1}{1-x}\)+ \(\frac{2x}{x^2-1}\)
d) \(\frac{1}{xy-x^2}\)- \(\frac{1}{y^2-xy}\)
bài 1: Thực hiện các phép tính
a.\(\frac{4x-1}{3x^2y}-\frac{7x-2}{3x^2y}\)
b.\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
c.\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d.\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
e. \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
f..\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
g. \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
h.\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
i.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
bài 1 : thực hiện các phép tính
a. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}\)
b.\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
c.\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d.\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
e.\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
f.\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
g.\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
h.\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
i.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
Tìm x, y, z biết:
a, \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}v\)à x+y=-24
b, \(\frac{x}{7}=\frac{y}{6}=\frac{z}{5}\)và 3z-2y=20
c, \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và x+2y-3z=-20
d, \(\frac{x}{2}=\frac{y}{3};\frac{y}{8}=\frac{z}{10}\)và x+y-z=20
e, 3x=2y;\(\frac{y}{6}=\frac{z}{7}\)và x+y-z=30
f, \(\frac{x}{2}=\frac{y}{3}\)và xy= 5400
Mấy bài còn lại tương tự nhé cậu
1 cho x,y,z là các số dương thỏa mãn \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\)
CM: \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)
2 Giải hệ pt
\(\left\{{}\begin{matrix}x^2+y^2-xy=5\\x^3+y^3=5x+15y\end{matrix}\right.\)
Bài 1:
Đặt \(\left(x+y;y+z;z+x\right)=\left(a;b;c\right)\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
\(P=\frac{1}{2a+b+c}+\frac{1}{a+b+2c}+\frac{1}{a+2b+c}\)
\(P=\frac{1}{a+a+b+c}+\frac{1}{a+b+c+c}+\frac{1}{a+b+b+c}\)
\(\Rightarrow P\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{2}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{6}{4}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\) hay \(x=y=z=\frac{1}{4}\)
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=5\\\left(x+y\right)\left(x^2+y^2-xy\right)=5x+15y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2-xy=5\\5\left(x+y\right)=5x+15y\end{matrix}\right.\)
\(\Rightarrow10y=0\Rightarrow y=0\)
Thay vào pt đầu: \(x^2=5\Rightarrow x=\pm\sqrt{5}\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(\sqrt{5};0\right);\left(-\sqrt{5};0\right)\)