Câu 1: \(\lim_{x\to2}\frac{2x^2-8}{3x-6}\)
\(=\lim_{x\to2}\frac{2\left(x-2\right)\left(x+2\right)}{3\left(x-2\right)}=\lim_{x\to2}\frac{2\left(x+2\right)}{3}=\frac{2\cdot\left(2+2\right)}{3}=\frac83\)
=>Chọn D
Câu 2: \(\lim_{x\to3}\frac{2x^2-5x-3}{3-x}\)
\(=\lim_{x\to3}\frac{2x^2-6x+x-3}{-\left(x-3\right)}=\lim_{x\to3}\frac{\left(x-3\right)\left(2x+1\right)}{-\left(x-3\right)}=\lim_{x\to3}\left\lbrack-\left(2x+1\right)\right\rbrack=-\left\lbrack2\cdot3+1\right\rbrack=-7\)
=>Chọn A
Câu 3: B
Câu 4: \(\lim_{x\to1}\frac{x^2-2a}{x-1}=b\)
=>\(1^2-2a=0\)
=>2a=1
=>a=1/2
\(\lim_{x\to1}\frac{x^2-2a}{x-1}=b\)
=>\(\lim_{x\to1}\frac{x^2-2\cdot\frac12}{x-1}=b\)
=>\(\lim_{x\to1}\frac{x^2-1}{x-1}=b\)
=>\(b=\lim_{x\to1}x+1=1+1=2\)
ab=1/2*2=1
=>Chọn D
Câu 5: \(\lim_{x\to1}\frac{x^{m}-1}{x-1}=7\)
=>m=7
\(\lim_{x\to1}\frac{x^{m}-x^{n}}{x-1}=2\)
=>\(\lim_{x\to1}\frac{\left(x^{m}-1\right)-\left(x^{n}-1\right)}{x-1}=2\)
=>\(\lim_{x\to1}\left\lbrack\frac{x^{m}-1}{x-1}-\frac{x^{n}-1}{x-1}\right\rbrack=2\)
=>m-n=2
=>7-n=2
=>n=5
\(m^2+n^2=7^2+5^2=49+25=74\)
=>Chọn A
