2a/3b+3b/4c+4c/5d+5d/2a biet 2a/3b=3b/4c=4c/5d=5d/2a

=>\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=\frac{2ax3bx4cx5d}{2ax3bx4cx5b}=1\)

=>1--1,4=-4

=> -4 hoac 4

b: T=(-1)+(-1)+...+(-1)

=-1011

a)x=-2

b) y=10

c)x=-5

d) x=-12

\(a,\dfrac{x}{3}=\dfrac{6}{-9}\\ \Rightarrow x=-\dfrac{2}{3}.3\\ \Rightarrow x=-2\\ b,\dfrac{4}{y}=\dfrac{-2}{-5}\\ \Rightarrow y=4:\dfrac{2}{5}\\ \Rightarrow y=10\\ c,\dfrac{-2}{3}=\dfrac{x-1}{6}\\ \Rightarrow3x-3=-12\\ \Rightarrow3x=-9\\ \Rightarrow x=-3\\ d,\dfrac{3}{x}=\dfrac{6}{-24}\\ \Rightarrow x=3:-\dfrac{1}{4}\\ \Rightarrow x=-12\)

a)x=-2

b) y=10

c)x=-5

d) x=-12 .

\(a,\dfrac{3}{4}=\dfrac{15}{20}\\ \dfrac{7}{10}=\dfrac{14}{20}\\ b,\dfrac{-5}{14}=\dfrac{-55}{154}\\ \dfrac{9}{22}=\dfrac{63}{154}\\ \dfrac{-4}{7}=\dfrac{-36}{63}\\ \dfrac{8}{9}=\dfrac{56}{63}\\ \dfrac{-10}{21}=\dfrac{-30}{63}\)

a: Để 5/n-1 là số nguyên thì \(n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

b: Để n+8/n+1 là số nguyên thì \(n+1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{0;-2;6;-8\right\}\)

Để 5/n-1 nhận giá trị là số nguyên thì: 

n+1 thuộc Ư(5)= {-1;1;-5;5}

Lập bảng:

=> n thuộc {0;-2;-6;4} thì n sẽ nhận giá trị là số nguyên

câu b làm tương tượng nhưng lấy n+1 thuộc Ư(7)

a: Để 5/n-1 là số nguyên thì 

hay 

b: Để n+8/n+1 là số nguyên thì 

hay 

Bài 4:

Gọi số học sinh là x

Theo đề, ta có: \(x\in BC\left(15;18;20\right)\)

hay x=540

540

HT

a: \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot6}{2}=12\\y=\dfrac{3\cdot\left(-2\right)}{6}=-1\end{matrix}\right.\)

b: =>-5<x<=4

hay \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

2,

\(a,\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\\ =1-\dfrac{1}{99}\\ =\dfrac{98}{99}\)

\(b,\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{40.43}\\ =\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\right):3\\ =\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\right):3\\ =\left(1-\dfrac{1}{43}\right):3\\ =\dfrac{42}{43}:3=\dfrac{14}{43}\)

\(1,\dfrac{a}{n\left(n+a\right)}=\dfrac{\left(a+n\right)-n}{n\left(a+n\right)}=\dfrac{a+n}{n\left(a+n\right)}-\dfrac{n}{n\left(a+n\right)}=\dfrac{1}{n}-\dfrac{1}{a+n}\)

\(M=x^4-4x+7=\left(x^2-4x+4\right)+3=\left(x-2\right)^2+3\ge3\)

\(minM=3\Leftrightarrow x=2\)

\(P=x^2-6x+y^2-2y+12=\left(x^2-6x+9\right)+\left(y^2-2y+1\right)+2=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

\(minP=2\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)