Toán Lớp 8: a/4x + 3 + 5x – 3
9x 9x
b/x^2 + y^2 + 2xy
x + y x + y x + y
c/3x + 5 + x – 25
x^2 – 5 x 5x – 25
d/ 1 – 1
xy – x^2 y^2 – xy
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a)\dfrac{{4x + 3}}{{9x}} + \dfrac{{5x – 3}}{{9x}}\\
= \dfrac{{4x + 3 + 5x – 3}}{{9x}}\\
= \dfrac{{9x}}{{9x}}\\
= 1\\
b)\dfrac{{{x^2}}}{{x + y}} + \dfrac{{{y^2}}}{{x + y}} + \dfrac{{2xy}}{{x + y}}\\
= \dfrac{{{x^2} + {y^2} + 2xy}}{{x + y}}\\
= \dfrac{{{{\left( {x + y} \right)}^2}}}{{x + y}}\\
= x + y\\
c)\dfrac{{3x + 5}}{{{x^2} – 5x}} + \dfrac{{x – 25}}{{5x – 25}}\\
= \dfrac{{3x + 5}}{{x\left( {x – 5} \right)}} + \dfrac{{x – 25}}{{5\left( {x – 5} \right)}}\\
= \dfrac{{\left( {3x + 5} \right).5 + x.\left( {x – 25} \right)}}{{5x\left( {x – 5} \right)}}\\
= \dfrac{{15x + 25 + {x^2} – 25x}}{{5x\left( {x – 5} \right)}}\\
= \dfrac{{{x^2} – 10x + 25}}{{5x\left( {x – 5} \right)}}\\
= \dfrac{{{{\left( {x – 5} \right)}^2}}}{{5x\left( {x – 5} \right)}}\\
= \dfrac{{x – 5}}{{5x}}\\
d)\dfrac{1}{{xy – {x^2}}} – \dfrac{1}{{{y^2} – xy}}\\
= \dfrac{1}{{x\left( {y – x} \right)}} – \dfrac{1}{{y\left( {y – x} \right)}}\\
= \dfrac{{y – x}}{{xy\left( {y – x} \right)}}\\
= \dfrac{1}{{xy}}
\end{array}$