tính
\(\left(\frac{5y+1}{y^2-5y}+\frac{5y-1}{y+5y}\right).\frac{y^2-25}{y^2+1}\)
Những câu hỏi liên quan
Cho \(\frac{x}{5y}=\frac{y}{2x+y}=\frac{10-5y}{x}\) với x,y khác 0, y khác 2x. Tính \(\left(\frac{x}{y}\right)^2+\left(x-5y\right)^{2015}\)
frac{1}{x}+frac{1}{y}frac{1}{5}Leftrightarrowfrac{x+y}{xy}frac{1}{5}Leftrightarrow5x+5yxyLeftrightarrow5x+5y-xy0Leftrightarrow xleft(5-yright)+5y0Leftrightarrow xleft(5-yright)+5y-25-25Leftrightarrow xleft(5-yright)-5left(5-yright)-25Leftrightarrowleft(5-yright)left(x-5right)-25nốt :V
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\(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{1}{5}\)
\(\Leftrightarrow5x+5y=xy\)
\(\Leftrightarrow5x+5y-xy=0\)
\(\Leftrightarrow x\left(5-y\right)+5y=0\)
\(\Leftrightarrow x\left(5-y\right)+5y-25=-25\)
\(\Leftrightarrow x\left(5-y\right)-5\left(5-y\right)=-25\)
\(\Leftrightarrow\left(5-y\right)\left(x-5\right)=-25\)
nốt :V
Thanks !!!!!!!!!!!!!!!
Rút gọn rồi tính: \(B=\frac{\left(x+5y\right)\left(x-5y\right)}{x^2+y^2}\left(\frac{5x-y}{x^2+5xy}+\frac{5x+y}{x^2-5xy}\right)\)
Bài 1
a)\(\frac{1}{x}+\frac{2}{x\left(x-2\right)}=\frac{x+2}{x-2}\)
b)\(\frac{y+5}{y^2-5y}-\frac{y-5}{2y^2+10y}=\frac{y+25}{2y^2-50}\)
8 tháng 12 2019 lúc 16:44
Giải hpt : a) left{{}begin{matrix}left(x^2+y^2right)left(x+y+1right)25left(y+1right)x^2+xy+2y^2+x-8y9end{matrix}right. b) left{{}begin{matrix}x^2+y^2+6xy-frac{1}{left(x-yright)^2}+frac{9}{8}02y-frac{1}{x-y}+frac{5}{4}0end{matrix}right.
c) left{{}begin{matrix}frac{x}{x^2-y}+frac{5y}{x+y^2}45x+y+frac{x^2-5y^2}{xy}5end{matrix}right. d) left{{}begin{matrix}3xy+y+121x9x^2y^2+3xy+1117x^2end{matrix}right.
e) left{{}begin{matrix}xleft(x^2-y^2right)+x^21sqrt{left(x-y^2right)^3}7...
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Giải hpt : a) \(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}x^2+y^2+6xy-\frac{1}{\left(x-y\right)^2}+\frac{9}{8}=0\\2y-\frac{1}{x-y}+\frac{5}{4}=0\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{x}{x^2-y}+\frac{5y}{x+y^2}=4\\5x+y+\frac{x^2-5y^2}{xy}=5\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}3xy+y+1=21x\\9x^2y^2+3xy+1=117x^2\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=1\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)
e) Sửa đề: \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=2\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)
PT(1) \(\Leftrightarrow x^3+x\left(x-y^2\right)=\sqrt{\left(x-y^2\right)^3}\)
Đặt \(\sqrt{x-y^2}=a.\text{Thay vào, ta có: }x^3+xa^2-2a^3=0\)
Làm tiếp như ở Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học toán với OnlineMath
Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira, Nguyễn Thị Ngọc Thơ, Nguyễn Lê Phước Thịnh, Quân Tạ Minh, An Võ (leo), @tth_new
e nhiều bài quá giải k kịp mn giúp e vs ạ!cần gấp lắm ạ
thanks nhiều!
\(\hept{\begin{cases}4\sqrt{1+\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}}+5y=\left(\sqrt{y}+2\sqrt{y+1}\right)^2\\4\sqrt{1+\frac{1}{y^2}+\frac{1}{\left(y+1\right)^2}}+5x=\left(\sqrt{x}+2\sqrt{x+1}\right)^2\end{cases}}\)
GIÚP MÌNH VỚI!Rút gọn phân thức:a) frac{2x+2y+5x+5y}{2x+2y-5x-5y}b) frac{15xleft(x+yright)^3}{5yleft(x+yright)^2}c) frac{5left(x-yright)-3left(y-xright)}{10left(x-yright)}d) frac{3left(x-yright)left(x-zright)^2}{6left(x-yright)left(x-zright)}h) frac{3xleft(1-xright)}{2left(x-1right)}j) frac{6x^2y^2}{8xy^5}
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GIÚP MÌNH VỚI!
Rút gọn phân thức:
a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}\)
b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\)
c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\)
d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}\)
j) \(\frac{6x^2y^2}{8xy^5}\)
a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)
b)\(=\frac{3x\left(x+y\right)}{y}\)
c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)
b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)
c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)
h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)
j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)
Câu b) bạn xem lại nhé.
Học tốt ^3^
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{3}{4}\\\frac{1}{6x}+\frac{1}{5y}=\frac{2}{15}\end{matrix}\right.\)
ĐK:\(x,y\ne0\). Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\).
Ta có hệ phương trình: \(\left\{ \begin{array}{l} a + b = \dfrac{2}{4}\\ \dfrac{1}{6}.a + \dfrac{1}{5}.b = \dfrac{2}{{15}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \dfrac{1}{2}\\ b = \dfrac{1}{4} \end{array} \right.\)
\(\Rightarrow \left\{ \begin{array}{l} \dfrac{1}{x} = \dfrac{1}{2}\\ \dfrac{1}{y} = \dfrac{1}{4} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 4 \end{array} \right.\left( {tm} \right)\)
ĐKXĐ: \(x,y\ne0\)
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b\)
Hệ trở thành: \(\left\{{}\begin{matrix}a+b=\frac{3}{4}\\\frac{a}{6}+\frac{b}{5}=\frac{2}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+4b=3\\5a+6b=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}20a+20b=15\\20a+24b=16\end{matrix}\right.\Rightarrow4b=1\Leftrightarrow b=\frac{1}{4}\);\(a=\frac{1}{2}\)
suy ra \(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
giải hệ phương trình: \(\hept{\begin{cases}\frac{\left(x-y\right)^2-1}{xy}-\frac{2\left(x+y-1\right)}{x+y}=-4\\4x^2+5y+\sqrt{x+y-1}+6\sqrt{x}=13\end{cases}}\)