Toán Lớp 8: 1.Trừ các phân thức:
a. 2x+8/x^2-4x+4 – 7/x-2
b. x – xy/x+y – x^3/(x+y)(x-y)
c. a+b/a^2-ab+b^2 – 1/a+b
d. a – 2 + 4a/a+2 – a^3+b/a^2-2a
2. Chứng minh rằng: 3a^2+3/a^3-1 – a–1/a^2+a+1 + 2/1-a=0
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Toán Lớp 8: 1.Trừ các phân thức:
a. 2x+8/x^2-4x+4 – 7/x-2
b. x – xy/x+y – x^3/(x+y)(x-y)
c. a+b/a^2-ab+b^2 – 1/a+b
d. a – 2 + 4a/a+2 – a^3+b/a^2-2a
2. Chứng minh rằng: 3a^2+3/a^3-1 – a–1/a^2+a+1 + 2/1-a=0
Comments ( 1 )
a)\dfrac{{2x + 8}}{{{x^2} – 4x + 4}} – \dfrac{7}{{x – 2}}\\
= \dfrac{{2x + 8}}{{{{\left( {x – 2} \right)}^2}}} – \dfrac{7}{{x – 2}}\\
= \dfrac{{2x + 8 – 7\left( {x – 2} \right)}}{{{{\left( {x – 2} \right)}^2}}}\\
= \dfrac{{22 – 5x}}{{{{\left( {x – 2} \right)}^2}}}\\
c)\dfrac{{a + b}}{{{a^2} – ab + {b^2}}} – \dfrac{1}{{a + b}}\\
= \dfrac{{{{\left( {a + b} \right)}^2} – {a^2} + ab – {b^2}}}{{\left( {a + b} \right)\left( {{a^2} – ab + {b^2}} \right)}}\\
= \dfrac{{3ab}}{{{a^3} + {b^3}}}\\
d)a – 2 + \dfrac{{4a}}{{a – 2}} – \dfrac{{{a^3} + b}}{{{a^2} – 2a}}\\
= \dfrac{{\left( {a – 2} \right).a\left( {a – 2} \right) + 4a.a – {a^3} – b}}{{a\left( {a – 2} \right)}}\\
= \dfrac{{{a^3} – 4{a^2} + 4a + 4{a^2} – {a^3} – b}}{{a\left( {a – 2} \right)}}\\
= \dfrac{{4a – b}}{{a\left( {a – 2} \right)}}\\
2)\dfrac{{3{a^2} + 3}}{{{a^3} – 1}} – \dfrac{{a – 1}}{{{a^2} + a + 1}} + \dfrac{2}{{1 – a}}\\
= \dfrac{{3{a^2} + 3}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}} – \dfrac{{a – 1}}{{{a^2} + a + 1}} – \dfrac{2}{{a – 1}}\\
= \dfrac{{3{a^2} + 3 – \left( {a – 1} \right)\left( {a – 1} \right) – 2\left( {{a^2} + a + 1} \right)}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{3{a^2} + 3 – {a^2} + 2a – 1 – 2{a^2} – 2a – 2}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{0}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= 0
\end{array}$
b)x – \dfrac{{xy}}{{x + y}} – \dfrac{{{x^3}}}{{\left( {x + y} \right)\left( {x – y} \right)}}\\
= \dfrac{{x\left( {x – y} \right)\left( {x + y} \right) – xy\left( {x – y} \right) – {x^3}}}{{\left( {x + y} \right)\left( {x – y} \right)}}\\
= \dfrac{{x\left( {{x^2} – {y^2}} \right) – {x^2}y + x{y^2} – {x^3}}}{{\left( {x + y} \right)\left( {x – y} \right)}}\\
= \dfrac{{{x^3} – x{y^2} – {x^2}y + x{y^2} – {x^3}}}{{\left( {x + y} \right)\left( {x – y} \right)}}\\
= \dfrac{{ – {x^2}y}}{{{x^2} – {y^2}}}
\end{array}$