Quy đồng
\(\dfrac{x^2+xy}{\left(x+y\right)^2};\dfrac{y^2-xy}{\left(x-y\right)^2};\dfrac{2xy}{x^2-y^2}\)
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QUy đồng
\(\dfrac{x^2+xy}{\left(x+y\right)^2};\dfrac{y^2-xy}{\left(x-y\right)^2};\dfrac{2xy}{x^2-y^2}\)
quy đồng mẫu thức của các phân tử
a, \(\dfrac{2}{5x^3y};\dfrac{5}{xy^2}\)
b, \(\dfrac{3x}{x-5};\dfrac{-2}{3\left(x-5\right)}\)
c, \(\dfrac{1}{\left(x+3\right)\left(x-3\right)};\dfrac{5x}{x^2-9}\)
Quy đồng mẫu các phân thức sau:
a)\(\dfrac{x}{x-y}\); \(\dfrac{y}{\left(x-y\right)^2}\) ; \(\dfrac{1}{\left(y-x\right)^3}\)
b) \(\dfrac{1}{2x+4};\dfrac{x}{2x-4};\dfrac{3}{4-x^2}\)
Quy đồng mẫu các phân thức sau:
a)\(\dfrac{x}{x-y};\dfrac{y}{\left(x-y\right)^2};\dfrac{1}{\left(y-x\right)^3}\)
b) \(\dfrac{1}{2x+4};\dfrac{x}{2x-4};\dfrac{3}{4-x^2}\)
Quy đồng mẫu thức các phân thức sau:
\(\dfrac{x+y}{x^{2^{ }}.(y+z)}\); \(\dfrac{y+z}{y^2.\left(z+x\right)}\); \(\dfrac{z+x}{z^2.\left(x+y\right)}\)
\(\dfrac{5x}{x^2+5x+6}\); \(\dfrac{2x+3}{x^2+7x+10}\); -5
Bài 2. Cho A=\(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\) :\([\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{1}{xy+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)]\)
Bạn cần làm gì với biểu thức này?
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Quy đồng mẫu thức của các phân thức
1. \(\dfrac{x-y}{2x^2-4xy+2y^2};\dfrac{x+y}{2x^2+4xy+2y^2};\dfrac{1}{y^2-x^2}\)
2. \(\dfrac{1}{x^2+8x+15};\dfrac{1}{x^2+6x+9}\)
3. \(\dfrac{1}{\left(a-b\right)\left(b-c\right)};\dfrac{1}{\left(c-b\right)\left(c-a\right)};\dfrac{1}{\left(b-a\right)\left(a-c\right)}\)
1: \(MTC=2\left(x-y\right)\left(x+y\right)\)
\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x+y}{2x^2+4xy+2y^2}\)
\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)
\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)
2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
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10 tháng 12 2023 lúc 15:10
y=\(2x-1\left(d_1\right)\)
y=\(\left(2n-1\right)x+\dfrac{3}{2}\left(d_2\right)\)
y=\(-x+3\left(d_3\right)\)
Tìm n đồng quy
Tọa độ giao điểm của (d1) và (d3) là:
\(\left\{{}\begin{matrix}2x-1=-x+3\\y=-x+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=4\\y=-x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{4}{3}+3=\dfrac{5}{3}\end{matrix}\right.\)
Thay x=4/3 và y=5/3 vào (d2), ta được:
\(\dfrac{4}{3}\left(2n-1\right)+\dfrac{3}{2}=\dfrac{5}{3}\)
=>\(\dfrac{8}{3}n-\dfrac{4}{3}+\dfrac{3}{2}=\dfrac{5}{3}\)
=>\(\dfrac{8}{3}n=\dfrac{5}{3}+\dfrac{4}{3}-\dfrac{3}{2}=\dfrac{3}{2}\)
=>\(n=\dfrac{3}{2}:\dfrac{8}{3}=\dfrac{3}{2}\cdot\dfrac{3}{8}=\dfrac{9}{16}\)
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F = \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]\)
\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
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