\(\Leftrightarrow x\left(x-1\right)+7⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{2;0;8;-6\right\}\)

`[2-x]/x >= 1`

`<=>[2-x-x]/x >= 0`

`<=>[2-2x]/x >= 0`

`<=>0 < x <= 1`

     `->\bb B`

B

B

\(\dfrac{19}{6}\)

=11/2

\(\dfrac{8}{15}:\dfrac{2}{11}+\dfrac{7}{15}:\dfrac{2}{1}=\dfrac{8}{15}\times\dfrac{11}{2}+\dfrac{7}{15}\times\dfrac{1}{2}=\dfrac{88}{30}+\dfrac{7}{30}=\dfrac{95}{30}=\dfrac{19}{6}\)

Cho g(x) = 0

x + 1 = 0

x = -1

Để f(x) chia hết cho g(x) thì x = -1 cũng là nghiệm của f(x)

Hay f(1) = 0

3.1² + 2.1² - 7.1 - m + 2 = 0

-2 - m + 2 = 0

m = 0

Vậy m = 0 thì f(x) chia hết cho g(x)

Giải chi tiết của em đây :

F(x) = 3x² + 2x² - 7x - m + 2 

F(x) \(⋮\) x + 1 \(\Leftrightarrow\) F(x) \(⋮\) x - (-1)

Theo bezout ta có : F(x) \(⋮\) x - (-1) \(\Leftrightarrow\) F(-1) = 0

\(\Leftrightarrow\) 3(-1)² + 2(-1)² - 7.(-1) - m + 2 = 0

    3 + 2 + 7 - m + 2 =0

              14 - m = 0

                     m = 14

Kết luận với m = 14 thì F(x) chia hết cho x + 1 

B

B. y=-3x-6

B

GIÚP MÌNH VỚI

Chọn A

B

\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{4x}{3x-3}\\ =\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}.\dfrac{3x-3}{4x}\\ =\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\dfrac{3\left(x-1\right)}{4x}\\ =\dfrac{4x.3\left(x-1\right)}{4x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{3}{x+1}\)

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

hay \(n\in\left\{2;0;4;-2\right\}\)

Ta có: \(\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{9}\right)>\dfrac{1}{9}.6=\dfrac{6}{9}>\dfrac{1}{2}\)  (1)

\(\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)>\dfrac{1}{19}.10=\dfrac{10}{19}>\dfrac{1}{2}\)  (2)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>\left(1\right)+\left(2\right)\)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>1\left(đpcm\right)\)

1/4+1/5+1/6+...+1/19=(1/4+1/5+...+1/11)+(1/12+1/13+...+1/19)>(1/11+1/11+...+1/11)+(1/19+1/19+...+1/19)=8/11+8/19=240/209>209/209=1

⇒B>1

Tham khảo:

Ta có :

\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\)

\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)\)

Ta thấy :

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}\)\(=\dfrac{1}{9}.5=\dfrac{5}{9}>\dfrac{1}{2}\)

\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}\)\(=\dfrac{1}{19}.5=\dfrac{5}{19}>\dfrac{1}{2}\)

⇒ \(B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}>1\)

⇒ \(B>1\) \(\left(đpcm\right)\)