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thi triển hđt ta đc a^3+b^3=(a+b)(a^2-ab+b^2)
suy ra a^2-ab+b^2=3
mà a+b=3
nên a=1;b=2
\(\left(a+b\right)^3-a^3-b^3=18\Leftrightarrow ab=2\)( vì a+b=3)
Đến đây tự làm tìm 2 só biết tổng, tích
\(a+b=3\Rightarrow b=3-a\)
\(a^3+b^3=9\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=9\Rightarrow3\left(a^2-ab+b^2\right)=9\Rightarrow a^2-ab+b^2=3\)(1)
\(a+b=3\Rightarrow\left(a+b\right)^2=3^2\Rightarrow a^2+2ab+b^2=9\)(2)
Trừ (2) cho (1), ta được: \(a^2+2ab+b^2-\left(a^2-ab+b^2\right)=9-3\)
\(\Rightarrow3ab=6\)
\(\Rightarrow ab=2\)
\(\Rightarrow a\left(3-a\right)=2\)
\(\Rightarrow3a-a^2=2\)
\(\Rightarrow a^2-3a=-2\)
\(\Rightarrow a^2-3a+2=0\)
\(\Rightarrow\left(a-1\right)\left(a-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}\Rightarrow\orbr{\begin{cases}b=3-1=2\\b=3-2=1\end{cases}}}\)
Vậy \(a=1,b=2\)hoặc \(a=2,b=1\)
Chúc bạn học tốt.
\(a,đk\left(B\right):x\ne\pm3\\ B=\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\\ =\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{x^2-9}\\ =\dfrac{3x+9+6x+x^2-3x}{x^2-9}\\ =\dfrac{x^2+6x+9}{x^2-9}\\ =\dfrac{\left(x+3\right)^2}{x^2-9}\\ =\dfrac{x+3}{x-3}\)
\(b,P=A.B\\ =\dfrac{x+1}{x+3}\times\dfrac{x+3}{x-3}\\ =\dfrac{x+1}{x-3}\)
\(c,\) Để P nguyên
\(\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}\)
=> \(x-3\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{-1;1;2;-2;4;-4\right\}\)
\(=>x=\left\{2;4;5;1;7;-1\right\}\)
1) A = \(\dfrac{2x-1}{x+3}\) = \(\dfrac{3}{2}\) (=) (2x-1).2 = 3.(x+3)
(=) 4x-2 =3x+9
(=) 4x-3x = 9+2
(=) x = 11 (tm)
2) Để \(\dfrac{A}{B}\)< \(^{x^2}\)+5 (=) \(\dfrac{2x-1}{x+3}\): \(\dfrac{2}{x^2-9}\) < \(x^2\)+5
(=) \(\dfrac{\left(2x-1\right)}{\left(x+3\right)}.\dfrac{\left(x-3\right)\left(x+3\right)}{2}\) < \(x^2\)+5
(=) \(\dfrac{\left(2x-1\right).\left(x-3\right)}{2}< x^2+5\)
(=) \(\dfrac{2x^2-6x-x+3}{2}\) < \(x^2\) +5
(=) \(2x^2\)- 7x + 3 < \(2x^2\)+ 10
(=) (\(2x^2\)-\(2x^2\)) - 7x < -3 +10
(=) -7x < 7
(=) x > -1