A=2/x-3/+5/y-1/+7
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Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)
2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)
3:
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)
4:
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)
3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)
\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)
\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=-\sqrt{a}\)
4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
\(\left(x-1\right)\left(y-5\right)=7\)
\(\left(x-1\right)\left(y-5\right)=7=1.7=7.1=-1.\left(-7\right)=-7.\left(-1\right)\)
x-1 | 1 | 7 | -1 | -7 |
y-5 | 7 | 1 | -7 | -1 |
x | 2 | 8 | 0 | -6 |
y | 12 | 6 | -2 | 4 |
vậy ...
mấy cái khác tương tự nha
\(\left(x+3\right)\left(xy+2\right)=3\)
\(\left(x+3\right)\left(xy+2\right)=3=1.3=3.1=-1.\left(-3\right)=-3.\left(-1\right)\)
\(th1\orbr{\begin{cases}x+3=1\\xy+2=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\-2y+2=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\-2y=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}}\)
\(th2\orbr{\begin{cases}x+3=3\\xy+2=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\0y+2=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\0y=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\y=0:1\left(ktm\right)\end{cases}}}\)
\(th3\orbr{\begin{cases}x+3=-1\\xy+2=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\-4y+2=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\-4y=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\y=-\frac{5}{4}\end{cases}}}\)
\(th4\orbr{\begin{cases}x+3=-3\\xy+2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\-6y+2=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\-6y=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\y=-\frac{1}{2}\end{cases}}}\)
vậy .......