Chọn B

\(\overline{ab}\times x=\overline{abab}\\ \overline{ab}\times x=\overline{ab00}+\overline{ab}\\ \overline{ab}\times x=\overline{ab}\times100+\overline{ab}\times1\\ \overline{ab}\times x=\overline{ab}\times\left(100+1\right)\\ \overline{ab}\times x=\overline{ab}\times101\\ x=101\)

\(B=\dfrac{2a+3}{a-2}=\dfrac{2\left(a-2\right)+7}{a-2}\\ =2+\dfrac{7}{a-2}\) (a nguyên, a khác 2)

Để B đạt gt nguyên thì: \(\dfrac{7}{a-2}\) cũng phải đạt gt nguyên

\(\Rightarrow7⋮\left(a-2\right)\)

\(\Rightarrow a-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\\ \Rightarrow a\in\left\{3;1;9;-5\right\}\left(TMDK\right)\)

\(x\times62+x\times48=4200\\ x\times\left(62+48\right)=4200\\ x\times110=4200\\ x=4200:110\\ x=\dfrac{420}{11}\)

X x 62 + X x 48 = 4200

X x (62 + 48) = 4200

X x 120 = 4200

X = 4200 : 120

X = 35

Vậy: ... 

P(x) có nghiệm khi P(x)=0

`=>x^2+5x+6=0`

`=>x^2+3x+2x+6=0`

`=>x(x+3)+2(x+3)=0`

`=>(x+3)(x+2)=0`

TH1: `x+3=0=>x=-3` (1)

TH2: `x+2=0=>x=-2` (2)

Từ (1) và (2) => `x=-3` và `x=-2` là 2 nghiệm của `P(x)`

\(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\\ =\left(\dfrac{-3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\\ =\dfrac{0}{5}+\dfrac{1}{3}\\ =0+\dfrac{1}{3}\\ =\dfrac{1}{3}\)

\(-\dfrac{3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\\ =\left(-\dfrac{3}{5}-\dfrac{4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\\ =\dfrac{-3-4+7}{5}+\dfrac{1}{3}\\ =\dfrac{0}{5}+\dfrac{1}{3}=0+\dfrac{1}{3}=\dfrac{1}{3}\)

Ta có: \(\widehat{E_1}\) và \(\widehat{F_2}\) là hai góc song le trong.

Nên \(\widehat{E_1}=\widehat{F_2}\)

Mà \(\widehat{E_1}=120^o\) suy ra \(\widehat{F_2}=120^o\)

Lại có, \(\widehat{F_1}\) và \(\widehat{F_2}\) là hai góc kề bù.

Nên \(\widehat{F_1}+\widehat{F_2}=180^o\)

Hay \(\widehat{F_1}+120^o=180^o\)

Suy ra: \(\widehat{F_1}=60^o\)

Sau hai lần lấy ra, số kg gạo còn lại trong bao chiếm số phần so với bao gạo ban đầu là: 

   \(1-\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

Ban đầu bao gạo có số kg gạo là:

  \(18:\dfrac{3}{8}=48\) (kg gạo)

    Đáp số: 48kg gạo

Thiếu dữ kiện rồi kìa, lần 2 lấy ra \(\dfrac{1}{8}\) bao gạo lúc đầu hay là lúc sau?

\(\dfrac{3}{2}+\dfrac{19}{6}+\dfrac{61}{12}+...+\dfrac{463}{42}\)

\(=1+3+5+...+11+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}\right)\)

Ta đặt biểu thức: 1 + 3 + 5 + ... + 11 = A

                             \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}\) = B

A = (11 + 1) x 6 : 2 = 36

B = Không rõ quy luật.

a) \(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{5}{4}-x=5\dfrac{1}{3}:0,8\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}\times\dfrac{5}{4}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=-\dfrac{65}{12}\)

b) \(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\times\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=\dfrac{-119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ \Rightarrow x=-\dfrac{1190}{9}\)

a) 

\(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{16}{3}:\left(\dfrac{5}{4}-x\right)=\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=\dfrac{-65}{12}\)

b) 

\(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\cdot\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=-\dfrac{119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ =-\dfrac{1190}{9}\)