a: Để A lớn nhất thì x-99=1

=>x=100

b: A=2012+555/1=2567

a, a x 6 = 3 x 6 = 18

b, a + b = 4 + 2 = 6

c, b + a = 2 + 4 = 6

d, a - b = 8 - 5 = 3

e, m x n = 5 x 9 = 45

a. 18

b. 6

c. 6

d. 3

e. 45

bạn cần câu nào?

a: \(\frac{A}{B}=\frac{2x^3+x^2-x+3}{x^2+x}\)

\(=\frac{2x^3+2x^2-x^2-x+3}{x^2+x}=\frac{2x\left(x^2+x\right)-\left(x^2+x\right)+3}{x^2+x}\)

\(=2x-1+\frac{3}{x^2+x}\)

b: Để A⋮B thì 3⋮\(x^2+x\)

mà \(x^2+x=x^2+x+\frac14-\frac14=\left(x+\frac12\right)^2-\frac14\ge-\frac14\forall x\)

nên \(x^2+x\in\left\lbrace1;3\right\rbrace\)

TH1: \(x^2+x=1\)

=>\(x^2+x+\frac14=\frac54\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt5}{2}\\ x+\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-1}{2}\left(loại\right)\\ x=\frac{-\sqrt5-1}{2}\left(loại\right)\end{array}\right.\)

TH2: \(x^2+x=3\)

=>\(x^2+x+\frac14=\frac{13}{4}\)

=>\(\left(x+\frac12\right)^2=\frac{13}{4}\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{\sqrt{13}}{2}\\ x+\frac12=-\frac{\sqrt{13}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{13}-1}{2}\left(loại\right)\\ x=\frac{-\sqrt{13}-1}{2}\left(loại\right)\end{array}\right.\)

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)

Có vài bước mình làm tắc á nha :>

a: Khi x=1 thì \(A=\dfrac{x-8}{x-3}=\dfrac{1-8}{1-3}=\dfrac{-7}{-2}=\dfrac{7}{2}\)

Khi x=2/11 thì \(A=\dfrac{\dfrac{2}{11}-8}{\dfrac{2}{11}-3}=\dfrac{-86}{11}:\dfrac{-31}{11}=\dfrac{86}{31}\)

b: Để A là số nguyên thì \(x-8⋮x-3\)

\(\Leftrightarrow x-3-5⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)