m = 3/2 = 1.5 >1

kich mk di

diem mk thap qua

thank you

Ta có: m=\(\frac{a}{c+b}+\frac{b}{c+a}+\frac{c}{a+b}\)

              = 1/2 <1

kb đc 0

2 câu đầu tôi làm đc

a) Ta có :

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)

\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)

vậy A > 1

b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)

Vậy B > \(\frac{1}{2}\)

Ta có : 

\(\frac{a}{b+c}>\frac{a}{a+b+c}\)

\(\frac{b}{c+a}>\frac{b}{a+b+c}\)

\(\frac{c}{a+b}>\frac{c}{a+b+c}\)

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

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

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

Chúc bạn học tốt !!! 

a/b+c > a/a+b+c           (1)

b/c+a > b/a+b+c           (2)

c/a+b > c/a+b+c           (3)

Lấy (1)+(2)+(3) ta có

a/b+c + b/c+a +c/a+b < 1

Ta có:

\(\frac{a}{b+c}>\frac{a}{a+b+c};\frac{b}{a+b+c}>\frac{b}{a+b+c};\frac{c}{a+b}>\frac{c}{a+b+c}\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+c+b}+\frac{c}{a+b+c}\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a+b+c}{a+b+c}\)

Ta thấy \(\frac{a+b+c}{a+b+c}=1\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>1\)

Vậy......

Cô nàng Vân Anh cũng hỏi câu này à?? Lạ nhé!!

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

\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)

\(\Rightarrow S=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{abc}{abc+c.abc+ca}\)

\(S=\frac{abc}{a.\left(bc+b+1\right)}+\frac{1}{1+b+bc}+\frac{abc}{ac.\left(bc+b+1\right)}\)

\(S=\frac{bc}{bc+b+1}+\frac{1}{1+b+bc}+\frac{b}{bc+b+1}\)

\(S=\frac{bc+b+1}{bc+b+1}\)

\(S=1\)

Điều kiện \(c\ge0\);\(a;b>0\)

Ta có: \(a>b\)

\(\Rightarrow ac\ge bc\)

\(\Rightarrow ac+ab\ge bc+ab\)

\(a.\left(b+c\right)\ge b.\left(c+a\right)\)

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

Tham khảo nhé~

cần gấp mai sẽ lam cho

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

\(=\left(1-\frac{b}{a+b}\right)+\left(1-\frac{c}{b+c}\right)+\left(1-\frac{a}{c+a}\right)\)

\(< 3-\left(\frac{b}{a+b+c}+\frac{c}{b+c+a}+\frac{a}{c+a+b}\right)=3-1=2\)

=>M < 2

Ta có: 

\(\frac{a}{a+b}<\frac{a+c}{a+b+c};\frac{b}{b+c}<\frac{b+a}{a+b+c};\frac{c}{c+a}<\frac{c+b}{a+b+c}\)

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

Vậy M < 2.

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)