a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)

\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)

\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)

\(\Leftrightarrow x=\frac{-33}{28}\)

b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+1\right)\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)

<=> \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}-\dfrac{x+1}{14}-\dfrac{x+1}{15}=0\)

<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

Do: \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{14}>0\) nên x + 1 = 0

Vậy x = -1

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)

=>\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}-\dfrac{x+1}{14}-\dfrac{x+1}{15}\)=> (x+1)(\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\))

=>x+1=0 (vì 1/10+1/11+1/12+1/13-1/14-1/15 khác 0)

=>x=-1

chúc bn học tốt ^^

Chọn C

Ta có: .

Số hạng tổng quát của khai triển là: . Hệ số của x k  trong khai triển là:  C n k

Hệ số của số hạng chứa  x 9 trong biểu thức P(x) là:

 .

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}+\frac{x+1}{15}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}-\frac{x+1}{15}=0\)

\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

\(\text{Vì }\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\text{ nên:}\)

\(x+1=0\)

\(x=-1\)

\(\text{Vậy x=-1}\)

Trương Chí Kiêng vừa ns thiếu để sao giờ giải đk rùi có siêu sao nhí 

x=-1 bạn nha

CHÚC BẠN HỌC GIỎI

báo cáo nhé?

 the only 

jjjjjji8hgjkoljjimkkpllkkkkkkllkkkkklkI-ol0loookkkkokpplplkoo

hack nè

1,307,674,368,000

x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

2: \(\Leftrightarrow\left(\dfrac{x+1}{12}-1\right)+\left(\dfrac{x+2}{13}-1\right)=\left(\dfrac{x+3}{14}-1\right)+\left(\dfrac{x+4}{15}-1\right)\)

=>x-11=0

=>x=11

3: \(\Leftrightarrow\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

=>x-100=0

=>x=100

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

=> x + 1 = 0 ( vì 1/10 + 1/11 + 1/12 - 1/13 - 1/14  khac 0 )  

=> x = -1