ĐK: a,b,c \(\ne\) 0

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Lại có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)

Với \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}\)

\(\Rightarrow\) \(\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Rightarrow\) \(\dfrac{b+c}{bc}=0\) \(\Rightarrow\) b + c = 0 (vì bc \(\ne\) 0 do a,b,c \(\ne\) 0)

\(\Rightarrow\) b = -c \(\Rightarrow\) b⁵ = (-c)⁵ \(\Rightarrow\) b⁵ + c⁵ = 0

Thay b⁵ + c⁵ = 0 vào M ta được:

M = (a¹⁹ + b¹⁹).(b⁵ + c⁵).(c²⁰⁰¹ + a²⁰⁰¹)

M = (a¹⁹ + b¹⁹).0.(c²⁰⁰¹ + a²⁰⁰¹)

M = 0 (đpcm)

Chúc bn học tốt!

Sửa đề:

\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)

\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)

\(=2001.\dfrac{1}{10}-3\)

\(=200,1-3=197,1\)

Vậy S = 197,1

\(S=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)

\(S=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(S=\frac{2001}{b+c}+\frac{2001}{c+a}+\frac{2001}{a+b}-3\)

\(S=2001\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

\(S=2001.\frac{1}{10}-3=\frac{1971}{10}\)

1/(a+b) + 1/(b+c) + 1/(c+a)=1/10

<=>(a+b+c)(1/a+b + 1/b+c + 1/c+a)=(a+b+c).1/10

<=>2001.(1/a+b + 1/b+c + 1/c+a)=200,1

<=>2001/a+b + 2001/b+c + 2001/c+a =200,1

<=>a+b+c/a+b + a+b+c/b+c + a+b+c/c+a=200,1

<=>a+b/a+b + c/a+b + b+c/b+c + a/b+c + c+a/c+a + b/c+a

<=>3+ c/a+b + a/b+c + b/c+a=200,1

<=>c/a+b + a/b+c + b/c+a=198,1

Bài 2: 

a: \(\dfrac{-1}{5}< 0< \dfrac{1}{1000}\)

b: \(\dfrac{267}{268}< 1< \dfrac{1347}{1343}\)

nên \(-\dfrac{267}{268}>-\dfrac{1347}{1343}\)

d: \(\dfrac{-181818}{313131}=\dfrac{-181818:10101}{313131:10101}=\dfrac{-18}{31}\)

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

\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)

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

\(=2001\cdot\frac{1}{10}-3=\frac{1971}{10}\)

Ta có :

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Mà \(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)

\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)

dân ta phải biết sử ta

cái gì ko biết lên tra google

Từ \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\)

Từ \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Leftrightarrow ad+ab< ba+bc\)

Vì ad < bc \(\Rightarrow\) ad + ab < ba + bc (đpcm)

Từ \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\Rightarrow\left(a+c\right)d< \left(b+d\right)c\)

\(\Leftrightarrow ad+cd< cb+cd\)

Vì ad > cb \(\Rightarrow\left(a+c\right)d< \left(b+d\right)c\) (đpcm)

các bài khác thì tự làm mình thấy bài nào cũng dễ mà

Em có cách khác!

\(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)

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

\(+\frac{a+b+c+d}{d+a+b}=50\)

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

\(+\frac{c}{d+a+b}+1=50\)

\(\Rightarrow\frac{d}{a+b+c}+\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}=46\)

Đề: \(a+b+c+d=2000\)

\(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)

Tính:

 \(S=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\)

Giải:

Có: \(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)

=> \(\frac{1}{2000-d}+\frac{1}{2000-a}+\frac{1}{2000-b}+\frac{1}{2000-c}=\frac{1}{40}\)

<=> \(\frac{2000}{2000-d}+\frac{2000}{2000-a}+\frac{2000}{2000-b}+\frac{2000}{2000-c}=\frac{2000}{40}\)

<=> \(1+\frac{d}{2000-d}+1+\frac{a}{2000-a}+1+\frac{b}{2000-b}+1+\frac{c}{2000-c}=50\)

<=> \(\frac{d}{a+b+c}+\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}=46\)

=> \(S=46\)