Bài 8:
a: Sửa đề: \(\frac{1}{x-y};\frac{1}{x^2-y^2};\frac{1}{x^3-y^3}\)
\(MTC=\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\)
\(\frac{1}{x-y}=\frac{1\cdot\left(x+y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}=\frac{\left(x+y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{1}{x^2-y^2}=\frac{1}{\left(x-y\right)\left(x+y\right)}=\frac{1\cdot\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}=\frac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{1}{x^3-y^3}=\frac{1}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{1\cdot\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{x+y}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
b: \(\frac{1}{a^2-4}=\frac{1}{\left(a-2\right)\left(a+2\right)}=\frac{1\cdot\left(a^2+2a+4\right)}{\left(a-2\right)\left(a+2\right)\left(a^2+2a+4\right)}=\frac{a^2+2a+4}{\left(a-2\right)\left(a+2\right)\left(a^2+2a+4\right)}\)
\(\frac{1}{a^3-8}=\frac{1}{\left(a-2\right)\left(a^2+2a+4\right)}=\frac{1\cdot\left(a+2\right)}{\left(a-2\right)\left(a+2\right)\left(a^2+2a+4\right)}=\frac{a+2}{\left(a-2\right)\left(a+2\right)\left(a^2+2a+4\right)}\)
\(\frac{1}{a+2}=\frac{1\cdot\left(a-2\right)\left(a^2+2a+4\right)}{\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)}=\frac{\left(a-2\right)\left(a^2+2a+4\right)}{\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)}\)
Bài 8:
a: \(A=\frac{a^2-6a+9}{3-a}\)
\(=\frac{\left(3-a\right)^2}{3-a}\)
=3-a
Bài 6:
\(M=\frac{x^3-4x}{\left(x+2\right)^2}\)
a: ĐKXĐ: x<>-2
b: ta có: \(M=\frac{x^3-4x}{\left(x+2\right)^2}\)
\(=\frac{x\left(x^2-4\right)}{\left(x+2\right)^2}\)
\(=\frac{x\left(x-2\right)\left(x+2\right)}{\left(x+2\right)^2}=\frac{x\left(x-2\right)}{x+2}\)
c: Khi x=98 thì \(M=\frac{98\cdot\left(98-2\right)}{98+2}=\frac{98\cdot96}{100}=94,08\)
em cần giải gấp bài 3 chi tiết mọi người giúp em với ạ. Làm bài dưới dạng phân số ạ em cần gấp
