\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2+y^2}+\dfrac{4x^3}{x^4+y^4}+\dfrac{8x^7}{x^8+y^8}\)
Những câu hỏi liên quan
Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
Đúng 1
Bình luận (0)
Thực hiện phép tính:
a) \(\dfrac{x}{2x-y}-\dfrac{2x-y}{4x-2y}\)
b)\(\dfrac{3x+1}{x^2-1}-\dfrac{x}{2x-2}\)
c) \(\dfrac{x-2}{x^2-4}-\dfrac{-8-x}{3x^2+6x}\)
d) \(\dfrac{2}{2x-3}-\dfrac{x}{2x+3}-\dfrac{2x+1}{9-4x^2}\)
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
Đúng 2
Bình luận (0)
Giải hệ phương trình:
a)left{{}begin{matrix}dfrac{3x+2}{x-1}-dfrac{3y-1}{y+2}0dfrac{2}{x-1}+dfrac{3}{y+2}1end{matrix}right.
b)left{{}begin{matrix}dfrac{4x-5}{x+1}+dfrac{2y-3}{y-5}8dfrac{3}{x+1}-dfrac{2}{y-5}-1end{matrix}right.
c)left{{}begin{matrix}dfrac{x+y-2}{x+1}+dfrac{3-x}{y+1}dfrac{5}{4}dfrac{3left(x+y-2right)}{x+1}-dfrac{5-x+2y}{y+1}dfrac{3}{4}end{matrix}right.
d)left{{}begin{matrix}dfrac{x-y+1}{x-3}+dfrac{x+1}{y-3}dfrac{-7}{2}dfrac{2left(x-y+1right)}{x-3}-dfrac{x+y-2}{y-3}-dfrac{9}{2}...
Đọc tiếp
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\)
\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\)
\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
----------------------------------------
\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
------------------------------------------
\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
Đúng 2
Bình luận (0)
Tìm y
\(\dfrac{2}{5}\) X y : \(\dfrac{7}{4}=\dfrac{7}{8}\)
2\(\dfrac{2}{5}\) : y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}-1\dfrac{2}{5}x\) y = 1\(\dfrac{1}{4}\)
\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)
y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)
y = \(\dfrac{245}{64}\)
Đúng 1
Bình luận (0)
2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{12}{5}\): y = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)
y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)
y = \(\dfrac{15}{13}\)
Đúng 1
Bình luận (0)
\(\dfrac{12}{5}\) - 1\(\dfrac{2}{5}\) \(\times\) y = 1\(\dfrac{1}{4}\)
\(\dfrac{12}{5}\) - \(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{5}{4}\)
\(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)
\(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{23}{20}\)
y = \(\dfrac{23}{20}\) : \(\dfrac{7}{5}\)
y = \(\dfrac{23}{28}\)
Đúng 1
Bình luận (0)
\(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\)
\(\dfrac{1}{x+2}+\dfrac{3}{x^2-4}+\dfrac{x-14}{\left(x^2+4x+4\right).\left(x-2\right)}\)
\(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\dfrac{1}{x+3}+\dfrac{1}{\left(x+3\right).\left(x+2\right)}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
Đúng 1
Bình luận (0)
8 tháng 6 2021 lúc 20:08
Tìm các số nguyên x,y biết:a)dfrac{6}{2x+1}dfrac{2}{7}b) dfrac{24}{7x-3}dfrac{-4}{25}c) dfrac{4}{x-6}dfrac{y}{24}dfrac{-12}{18}d) dfrac{-1}{5}ledfrac{x}{8}ledfrac{1}{4}e) dfrac{x+46}{20}xdfrac{2}{5}f) ydfrac{5}{y}dfrac{86}{y} ( xdfrac{2}{5};ydfrac{5}{y} là các hỗn số)
Đọc tiếp
Tìm các số nguyên x,y biết:
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
b) \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
d) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
f) \(y\dfrac{5}{y}=\dfrac{86}{y}\) ( \(x\dfrac{2}{5};y\dfrac{5}{y}\) là các hỗn số)
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)
Đúng 3
Bình luận (0)
BÀI 6 rút gọn phân thức
\(\dfrac{2(x+1)^{2}}{4x(x+1)}\)
\(\dfrac{(8-x)(-x-2)}{(x+2)^{2}}\)
\(\dfrac{2(x-y)}{y-x} \)
\(\dfrac{(x+2)^{2}}{2x+4}\)
a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)
\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)
\(=\dfrac{x+1}{2x}\)
b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)
\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(8-x\right)}{x+2}\)
\(=\dfrac{x-8}{x+2}\)
c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)
\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)
\(=-2\)
d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
Đúng 3
Bình luận (0)
ĐKXĐ: \(x\neq0;x\neq-1\)
\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)
$---$
ĐKXĐ: \(x\neq y\)
\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)
Đúng 1
Bình luận (0)
Thực hiện các phép tính với các phân thức sau:
a) dfrac{4a^2-3a+5}{a^3-1}-dfrac{1-2a}{a^2+a+1}-dfrac{6}{a-1}
b) dfrac{5}{a+1}-dfrac{10}{a-left(a^2+1right)}-dfrac{15}{a^3+1}
c) dfrac{2x}{x^2+2xy}+dfrac{y}{xy-2y^2}+dfrac{4}{x^2-4y^2}
d) dfrac{1}{x-y}+dfrac{3xy}{y^3-x^3}+dfrac{x-y}{x^2+xy+y^2}
e) dfrac{2x+y}{2x^2-xy}+dfrac{16x}{y^2-4x^2}+dfrac{2x-y}{2x^2+xy}
f) dfrac{1}{1-x}+dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{1+x^{16}}
Đọc tiếp
Thực hiện các phép tính với các phân thức sau:
a) \(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\)
b) \(\dfrac{5}{a+1}-\dfrac{10}{a-\left(a^2+1\right)}-\dfrac{15}{a^3+1}\)
c) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
d) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
e) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
f) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
Đúng 0
Bình luận (0)