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Ta có : Cy(x+1) : Cy+1(x) = 6:5
<=> 5(x-1)(y+1) = 6(x-y)(x-y+1) (1)
Lại có : Cy(x+1) : Cy-1(x) = 6:2
<=> x+1=3y (2)
Thay (2) vào (1) => 9y^2-27y=0
<=> y=3 hoặc y=0(loại)
=> x=8
a, A= 23 - x - x + 46 + 2x - 43
A= 26
b, B = 4 + 2x + 76 - x - x - 99
B = -19
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d) y' =\(\dfrac{\left(x^2+7x+3\right)'\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(x^2-3x\right)'}{\left(x^2-3x\right)^2}\)=\(\dfrac{\left(2x+7\right)\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(2x-3\right)}{\left(x^2-3x\right)^2}\)=\(\dfrac{-2x^2-6x+9}{\left(x^2-3x\right)^2}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
a: \(P=x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b: \(P=3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)\)
\(=\left(x+5\right)\left(3x-1\right)\)
c: \(P=-2x^2-7x-5\)
\(=-\left(2x^2+7x+5\right)\)
\(=-\left(2x^2+2x+5x+5\right)\)
\(=-\left[2x\left(x+1\right)+5\left(x+1\right)\right]\)
\(=-\left(x+1\right)\left(2x+5\right)\)