Toán Lớp 8: bài 3 Cho hai phân thức A=(x+5)^2-9/x^2+4x+4 vàB=x(x+2)^2+4x+8/x^3-8
a)rút gọn A và b
B)tính A+B và a-B
bài 4 cho biểu thức A=6x/x^2-9x+5x/x^2-3x+x/x^2+3x
a)rút gọn bt A
b)Tìm các giá trị nguyên của x để bt A có gía trị nguyê
c)Tìm x để A=-1/2
giúp mình nhaa
Leave a reply
Comments ( 1 )
3)a)\\
A = \dfrac{{{{\left( {x + 5} \right)}^2} – 9}}{{{x^2} + 4x + 4}}\\
= \dfrac{{\left( {x + 5 + 3} \right)\left( {x + 5 – 3} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{\left( {x + 8} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{x + 8}}{{x + 2}}\\
B = \dfrac{{x{{\left( {x + 2} \right)}^2} + 4x + 8}}{{{x^3} – 8}}\\
= \dfrac{{x{{\left( {x + 2} \right)}^2} + 4\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x\left( {x + 2} \right) + 4} \right)}}{{\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{x + 2}}{{x – 2}}\\
b)A + B\\
= \dfrac{{x + 8}}{{x + 2}} + \dfrac{{x + 2}}{{x – 2}}\\
= \dfrac{{\left( {x + 8} \right)\left( {x – 2} \right) + {{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
= \dfrac{{{x^2} + 6x – 16 + {x^2} + 4x + 4}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
= \dfrac{{2{x^2} + 10x – 12}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
A – B\\
= \dfrac{{x + 8}}{{x + 2}} – \dfrac{{x + 2}}{{x – 2}}\\
= \dfrac{{\left( {x + 8} \right)\left( {x – 2} \right) – {{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
= \dfrac{{{x^2} + 6x – 16 – {x^2} – 4x – 4}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
= \dfrac{{2x – 20}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
B4)\\
a)Dkxd:x \ne 0;x \ne 3;x \ne – 3\\
A = \dfrac{{6x}}{{{x^2} – 9x}} + \dfrac{{5x}}{{{x^2} – 3x}} + \dfrac{x}{{{x^2} + 3x}}\\
= \dfrac{6}{{x – 9}} + \dfrac{5}{{x – 3}} + \dfrac{1}{{x + 3}}\\
= \dfrac{{6\left( {x – 3} \right)\left( {x + 3} \right) + 5\left( {x – 9} \right)\left( {x + 3} \right) + \left( {x – 9} \right)\left( {x – 3} \right)}}{{\left( {x – 9} \right)\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{6{x^2} – 54 + 5{x^2} – 30x – 135 + {x^2} – 12x + 27}}{{\left( {x – 9} \right)\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{12{x^2} – 42x – 162}}{{\left( {x – 9} \right)\left( {x – 3} \right)\left( {x + 3} \right)}}
\end{array}$