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 

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

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

hay \(x\in\left\{2;0;8;-6\right\}\)

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 :

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

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

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

⇒ \(B>1\) \(\left(đpcm\right)\)

a) \(x=-\dfrac{3}{5}\times\dfrac{9}{7}=-\dfrac{27}{35}\)

b) \(x\left(0,4-\dfrac{1}{5}\right)=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)

a, \(x=-3,5.\dfrac{9}{7}=-\dfrac{9}{2}\)

b, \(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)

x=\(\dfrac{9}{7}x-3.5\)

x=\(\dfrac{-9}{5}\)

ta có: 5/x = 1/8 => x= 1/8 x 5 = 5/8

=> x= 8

duyệt đi

B

B. y=-3x-6

B

`@` `\text {Ans}`

`\downarrow`

Ta có: \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)

\(\Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)`=`\(\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)

`=`\(\dfrac{5z-25-3x+3-4y-12}{8}\)

`=`\(\dfrac{\left(5z-3x-4y\right)+\left(-25+3-12\right)}{8}\)

`=`\(\dfrac{50-34}{8}\)`=`\(\dfrac{16}{8}=2\)

`=>`\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=2\)

`=>`\(\left\{{}\begin{matrix}x=2\cdot2+1=5\\y=2\cdot4-3=5\\z=2\cdot6+5=17\end{matrix}\right.\)

Vậy, `x,y,z` lần lượt là `5; 5; 17.`

\(\dfrac{3}{x}=\dfrac{1}{y}=\dfrac{6}{2}\)

\(\Rightarrow\dfrac{3}{x}=\dfrac{1}{y}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)

x = 1

y = 1/3

x = 1 

y = 1/3.