Question

[Sailor Moon]

Câu 1: Tìm x

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

Câu 2: Tìm các số nguyên n để A có giá trị nguyên

a) \(A=\dfrac{3n+9}{n-4}\)

Câu 3: Cho hai số hữu tỉ \(\dfrac{a}{b}và\dfrac{c}{d}\) với mẫu dương, trong đó \(\dfrac{a}{b}< \dfrac{c}{d}CMR\)

\(\dfrac{a}{b}< \dfrac{a+c}{b+c}< \dfrac{c}{d}\)

help me

Answer 1

[Nanami Luchia]

Câu 1:

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

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

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

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\)

=> x = 0 - 1

=> x = -1

Answer 2

[Nanami Luchia]

Câu 2:

Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-3.4+9+12}{n-4}\)

\(=\dfrac{3.\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)

Để A có giá trị nguyên thì:

n - 4 \(\in\) Ư(21)

=> n - 4 \(\in\)

n4	3	-3	7	-7	-1	1	-21	21
n	7	1	11	-3	3	5	-17	25