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◤(¬‿¬)◥ (∪ ◡ ∪) (*^ -^*) (●*∩_∩*●)

◖♪_♪|◗ •(_)• !⑈ˆ~ˆ!⑈ ⋋ō_ō` ‹(•¿•)›

Ta có : \(\hept{\begin{cases}\frac{a}{a+b}>\frac{a}{a+b+c}\\\frac{b}{b+c}>\frac{b}{a+b+c}\\\frac{c}{c+a}>\frac{c}{a+b+c}\end{cases}}\)

\(\Rightarrow M>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

=> M > 1 (1)

Lại có : \(\hept{\begin{cases}\frac{a}{a+b}< \frac{a+c}{a+b+c}\\\frac{b}{b+c}< \frac{a+b}{a+b+c}\\\frac{c}{c+a}< \frac{b+c}{a+b+c}\end{cases}}\)

\(\Rightarrow M< \frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

=> M > 2 (2)

Từ (1) và (2)

=> 1 < M < 2

=> M không phải là số nguyên 

1- B

2-C

3-C

4-C

Choose the best answer(A, B, C or D)

1. Today, subjects music and arts are put into the school __________ in VietNam.

A. so                             B. too                             c. either                             D.however

2. My cousin - Hoa, is the same age __________ me

A. as                             B. of                             C. to                             D.

3. For many pepple, a good knowledge of music and arts is regarded as a __________ for every stufent

A. need                             B. needs                             C. necessity                             D. necessary

4. This year's musical festival is not __________ it was last year

A. worse                             B. as good as                             C. as well as                             D. different from

5. Since the 1970, the festival in Glastonbury has taken __________ almost every year and has grown in size

A. time                             B. place                             C. note                             D. noe

Ta có:

^A-^B=18°

^B-^C=18°

=>^A-^B+^B-^C=18°+18°

=>^A-^C=36°

=>^C=^A-36°

Vì ^A-^B=18°

=>^B=^A-18°

Xét ΔABC có: ^A+^B^C=180°(đl tổng ba góc tam giác) 

=>^A+^A-18°+^A-36°=180°

=>3^A=234°

=>^A=78°

=>^B=78°-18°=60°

=>^C=78°-36°=42°

Từ  \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\)

\(\Rightarrow\frac{a+b-c}{c}+2=\frac{a-b+c}{b}+2=\frac{-a+b+c}{a}+2\)

\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)

Nếu a + b + c = 0

=> a + b = - c

=> b + c = - a

=> c + a = - b

Khi đó \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=\frac{-a.\left(-b\right).\left(-c\right)}{abc}=-\frac{abc}{abc}=-1\)

Nếu \(a+b+c\ne0\)

\(\Rightarrow\frac{1}{c}=\frac{1}{b}=\frac{1}{a}\)

\(\Rightarrow a=b=c\)

Khi đó \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=\frac{2a.2b.2c}{abc}=\frac{8.abc}{abc}=8\)

Vậy nếu a + b + c = 0 thì \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=-1\) 

nếu a + b + c \(\ne\)0 thì  \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=8\)