\(B=-x^2-x+5=-\left(x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}-\frac{1}{4}-5\right)=-\left(x+\frac{1}{2}\right)^2+5\frac{1}{4}\le5\frac{1}{4}\)
vậy để b max thì \(-\left(x+\frac{1}{2}\right)^2max\) mà \(-\left(x+\frac{1}{2}\right)^2\le0\)nên suy ra \(-\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Đáp án B

Dùng máy tính bỏ túi tính  ∫ 0 1 x d x 5 x 2 + 4 = 1 5 ⇒ T = 1 2 + 5 2 = 26

`(2-2/5+6/5)-(3/15+2)+(7/3-2)`

`=2+4/5-3/15-2-2+7/3`

`=-2+4/5-1/5+7/3`

`=-2+3/5+2+1/3`

`=3/5+1/3=14/15`

`=>a=14,b=15`

`=>S=2.14-15=13`

ĐK:n≠-2

Gọi \(d=ƯCLN\left(n+3,n+2\right)\)

\(\Rightarrow n+3⋮d;n+2⋮d\\ \Rightarrow n+3-n-2⋮d\\ \Rightarrow1⋮d\\ \Rightarrow d=1\)

Vậy n+3 và n+2 nctn hay \(\dfrac{n+3}{n+2}\) tối giản

Với n=-2 trái vs ĐKXĐ nên A ko xác định

\(\frac{3}{4}+\frac{5}{8}-\frac{5}{6}\)

\(=\frac{3}{4}\times1+\frac{5}{8}\times1-\frac{5}{6}\times1\)

\(=\frac{3}{4}\times\frac{6}{6}+\frac{5}{8}\times\frac{3}{3}-\frac{5}{6}\times\frac{4}{4}\)

\(=\frac{3\times6}{4\times6}+\frac{5\times3}{8\times3}-\frac{5\times4}{6\times4}\)

\(=\frac{18}{24}+\frac{15}{24}-\frac{20}{24}\)

\(=\frac{18+15-20}{24}\)

\(=\frac{13}{24}\)

Vậy GTBT \(\frac{3}{4}+\frac{5}{8}-\frac{5}{6}\) là \(\frac{13}{24}\)

\(\frac{3}{4}+\frac{5}{8}-\frac{5}{6}\)

\(=\frac{3\times6+5\times3-5\times4}{24}\)

\(=\frac{18+15-20}{24}\)

\(=\frac{33-20}{24}\)

\(=\frac{13}{24}\)

Đáp án B

Đáp án là A