\(\left(\dfrac{x^2}{y^2}+\dfrac{x}{y}\right):\left(\dfrac{x}{y^2}-\dfrac{1}{y}+\dfrac{1}{x}\right)\)
giải dúp đi mn ui
tính đạo hàm
a) \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}\)
b) \(y=x+3+\dfrac{4}{x+3}\) giải phương trình y'=0
c) \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\) tính y'(-1)
d) \(y=x-2+\dfrac{9}{x-2}\) giải phương trình y'=0
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
Rút gọn biểu thức P=\(\sqrt{\dfrac{1}{x^2+y^2}+\dfrac{1}{\left(x+y\right)^2}+\sqrt{\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{\left(x^2+y^2\right)^2}}}\)
Mn giúp e với
Giải hpt
a)\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}+\dfrac{xy}{x+y}=\dfrac{5}{2}\\\dfrac{x-y}{xy}+\dfrac{xy}{x-y}=\dfrac{10}{3}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
a: ĐKXĐ: x<>-1 và y<>-1
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+2-2}{x+1}+\dfrac{y+1-1}{y+1}=2\\\dfrac{x+1-1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{2}{x+1}+1-\dfrac{1}{y+1}=2\\1-\dfrac{1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=2-3=-1\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=1+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=-1\\\dfrac{2}{x+1}-\dfrac{6}{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y-1}=3\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=-\dfrac{7}{3}\\\dfrac{1}{x+1}-3:\dfrac{-7}{3}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}+3\cdot\dfrac{3}{7}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}=2-\dfrac{9}{7}=\dfrac{5}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x+1=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: y<>0 và y<>-12
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y}-\dfrac{x}{y+12}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\cdot\dfrac{x}{y+12}=3\left(vôlý\right)\\\dfrac{x}{y}-\dfrac{x}{y+12}=1\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\varnothing\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)\)
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)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
Tính:
a) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
b) \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-x\right)\left(y-z\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
d) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
Giúp mình với!!! Mình cần gấp!!! 10 giờ sáng mai cần gấp nha !!!
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
GHPT: \(\left\{{}\begin{matrix}2x+\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{16}{3}\\2\left(x^2+y^2\right)+\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y\right)^2+\dfrac{1}{\left(x+y\right)^2}+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y+\dfrac{1}{x+y}\right)^2+\left(x-y+\dfrac{1}{x-y}\right)^2=\dfrac{136}{9}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)
Hệ cơ bản, chắc bạn tự giải quyết phần còn lại được
\(\left\{{}\begin{matrix}^{x^2}+\dfrac{1}{^{y^2}}+\dfrac{x}{y}=1\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\end{matrix}\right.\)
Giải Hệ Phương Trình
Đk: \(y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\end{matrix}\right.\)
\(\Rightarrow-\left(x+\dfrac{1}{y}\right)^2+\dfrac{x}{y}=\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)\)
\(\Leftrightarrow-\left(x+\dfrac{1}{y}\right)^2+2\left(x+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=0\\x+\dfrac{1}{y}=2\end{matrix}\right.\)
TH1: \(x+\dfrac{1}{y}=0\Leftrightarrow\dfrac{1}{y}=-x\) thay vào pt dưới ta được:
\(-x^2=-1\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y=-1\\x=-1\Rightarrow y=1\end{matrix}\right.\)
TH2: \(x+\dfrac{1}{y}=2\Leftrightarrow\dfrac{1}{y}=2-x\) thay vào pt dưới ta được:
\(\left(2-x\right)x-2.2=-1\)\(\Leftrightarrow x^2-2x+3=0\left(vn\right)\)
Vậy (x;y)=(-1;1);(1;-1)
gợi ý \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\left(1\right)\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\left(2\right)\end{matrix}\right.\)
Đem \(\left(1\right)+\left(2\right):\left(x+\dfrac{1}{y}\right)^2-2\left(x+\dfrac{1}{y}\right)=0\)
đến đây chắc bạn có thể tự làm được
a) \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)
b) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
c) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
Chứng minh đẳng thức:
a) \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}+\dfrac{z}{\left(y-z\right)\left(z-x\right)}+\dfrac{x}{\left(z-x\right)\left(x-y\right)=0}\)
b) \(\dfrac{x^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)=1}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)