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a) \(\dfrac{x+9}{x^2-9}\)-\(\dfrac{3}{x^2+3x}\) = \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{3}{x\left(x+3\right)}\)
= \(\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x+3}{x\left(x-3\right)}\)
\(\dfrac{1}{x^2-4}+\dfrac{2x}{x+2}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x}{x+2}=\dfrac{1+2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1+2x^2-4x}{\left(x+2\right)\left(x-2\right)}\)
trên bài mink đã ẩn đi bước quy đồng!!
\(\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{18}{\left(x-3\right)^2\left(x+3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}=\dfrac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{9-x^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-1}{x-3}\)
a: Xét tứ giác ADHE có
góc ADH=góc AEH=góc DAE=90 độ
nên ADHE là hình chữ nhật
b: \(HD=\sqrt{10^2-8^2}=6\left(cm\right)\)
\(S_{ADHE}=6\cdot8=48\left(cm^2\right)\)
c: Để ADHE là hình vuông thì AH là phân giác của góc BAC
=>góc B=45 độ
\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)
\(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x\left(x-3\right)}\)
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\(\dfrac{x+1}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{x+1}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{x^2+x-x+6}{2x\left(x+3\right)}=\dfrac{x^2+6}{2x\left(x+3\right)}\)
a: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
b: \(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
c: \(=\dfrac{6-7+x}{3\left(x-1\right)}=\dfrac{x-1}{3\left(x-1\right)}=\dfrac{1}{3}\)
d: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
a: Xét ΔABC có BM/BC=BD/BA
nên MD//AC
=>MM' vuông góc AB
=>M đối xứngM' qua AB
b: Xét tứ giác AMBM' có
D là trung điểm chung của AB và MM'
MA=MB
Do đó: AMBM' là hình thoi
\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)
= \(\dfrac{x+9}{\left(x-3\right).\left(x+3\right)}-\dfrac{3}{x.\left(x+3\right)}\)
=\(\dfrac{\left(x+9\right).x}{\left(x-3\right).\left(x+3\right).x}-\dfrac{3.\left(x-3\right)}{x.\left(x+3\right).\left(x-3\right)}\)
=\(\dfrac{x^2+9x}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3x-9}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2+9-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2-3x+18}{3\left(x-3\right)\left(x+3\right)}\)