\(1,\frac{x}{4}=\frac{y}{2}\&x+y=6\)
\(2,x:y:z=6:7:8\&x+y+z=21\)
\(3,4x=5y\&x-y=1\)
Những câu hỏi liên quan
\(Cho A=\frac{1}{(x+y)^3}(\frac{1}{x^4+y^4})\) ;\(B=\frac{2}{(x+y)^4}(\frac{1}{x^3}-\frac{1}{y^3})\) :C=\(\frac{2}{(x+y)^5}(\frac{1}{x^2}-\frac{1}{y^2})\) Tính A+B+C \)
Tìm x,y ϵ Z
a,frac{1}{x}frac{1}{6}+frac{y}{3}
b,frac{x}{6}-frac{1}{y}frac{1}{2}
c,frac{x}{4}-frac{1}{y}frac{3}{4}
d,frac{x}{8}-frac{2}{y}frac{3}{4}
e,frac{x}{4}-frac{2}{y}frac{3}{2}
g,frac{1}{x}-frac{1}{y}frac{1}{x}.frac{1}{y}
x≠y≠0
Đọc tiếp
Tìm x,y ϵ Z
a,\(\frac{1}{x}=\frac{1}{6}+\frac{y}{3}\)
b,\(\frac{x}{6}-\frac{1}{y}=\frac{1}{2}\)
c,\(\frac{x}{4}-\frac{1}{y}=\frac{3}{4}\)
d,\(\frac{x}{8}-\frac{2}{y}=\frac{3}{4}\)
e,\(\frac{x}{4}-\frac{2}{y}=\frac{3}{2}\)
g,\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)
x≠y≠0
\(Cho A=\frac{1}{(x+y)^3}(\frac{1}{x^4+y^4})\) ;\(B=\frac{2}{(x+y)^4}(\frac{1}{x^3}-\frac{1}{y^3})\) :C=\(\frac{2}{(x+y)^5}(\frac{1}{x^2}-\frac{1}{y^2})\)
Tính A+B+C
Thực hiện phép tính :a)frac{x^2}{left(x-yright)^2left(x+yright)}-frac{2xy^2}{x^4-2x^2y^2+y^4}+frac{y^2}{left(x^2-y^2right)left(x+yright)} b)frac{1}{x-1}-frac{1}{x+1}-frac{2}{x^2+1}-frac{4}{x^4+1}-frac{8}{x^{8+1}}-frac{16}{x^{16}+1}c)frac{1}{x^2+6x+9}+frac{1}{6x-x^2-9}+frac{x}{x^2-9}d)frac{a}{x^2+ax}+frac{a}{x^2+3ax+2a^2}+frac{a}{x^2+5ax+6a^2}+....+frac{a}{x^2+19ax+90a^2}+frac{1}{x+10a}
Đọc tiếp
Thực hiện phép tính :
a)\(\frac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\frac{2xy^2}{x^4-2x^2y^2+y^4}+\frac{y^2}{\left(x^2-y^2\right)\left(x+y\right)}\)
b)\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^{8+1}}-\frac{16}{x^{16}+1}\)
c)\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
d)\(\frac{a}{x^2+ax}+\frac{a}{x^2+3ax+2a^2}+\frac{a}{x^2+5ax+6a^2}+....+\frac{a}{x^2+19ax+90a^2}+\frac{1}{x+10a}\)
1/ Cho \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}\), \(z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}\) và \(x\ne1,x\ne-1\). Hãy tính z theo y
2/ Cho xy+yz+xz=1 và x,y,z khác 1,-1. Chứng minh rằng \(\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\)
cho \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}},z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}\)
biễu diễn z theo y,tính z khi \(y^3-2y^2+y-2=0\)
ta thấy \(\left(x^2+\frac{1}{x^2}\right)\left(x^2-\frac{1}{x^2}\right)=\left(x^4-\frac{1}{x^4}\right)\)
\(\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right)=\left(x^4+\frac{1}{x^4}\right)+2\)
suy ra \(y=\frac{\left(x^4+\frac{1}{x^4}\right)+2}{\left(x^4-\frac{1}{x^4}\right)}\)
<=> \(y=z+\frac{2}{\left(x^4-\frac{1}{x^4}\right)}\)
<=>\(z=\frac{2}{\left(x^4-\frac{1}{x^4}\right)}-y\)
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\(\left\{{}\begin{matrix}x^2+x+\frac{1}{y}\left(1+\frac{1}{y}\right)=4\\x^3+\frac{x}{y^2}+\frac{x^2}{y}+\frac{1}{y^3=4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+x+\frac{1}{y}\left(1+\frac{1}{y}\right)=4\\x^3+\frac{x}{y^2}+\frac{x^2}{y}+\frac{1}{y^3}=4\end{matrix}\right.\)
ĐKXĐ: ...
\(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+x+\frac{1}{y}=4\\x^2\left(x+\frac{1}{y}\right)+\frac{1}{y^2}\left(x+\frac{1}{y}\right)=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+x+\frac{1}{y}=4\\\left(x^2+\frac{1}{y^2}\right)\left(x+\frac{1}{y}\right)=4\end{matrix}\right.\)
Đặt \(\left(x^2+\frac{1}{y^2};x+\frac{1}{y}\right)=\left(u;v\right)\Rightarrow\left\{{}\begin{matrix}u+v=4\\uv=4\end{matrix}\right.\)
Theo Viet đảo, u và v là nghiệm:
\(t^2-4t+4=0\Rightarrow t=2\Rightarrow\left\{{}\begin{matrix}x^2+\frac{1}{y^2}=2\\x+\frac{1}{y}=2\end{matrix}\right.\)
Bạn tự giải nốt
hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
Đọc tiếp
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
8,Thực hiện phép tínha,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}}
Đọc tiếp
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,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)