ban oi mk dat cau hoi nay cac ban giup mk vs

1/2x + 3/5 . ( x- 2 ) = 3

Ta có :

\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\)

\(\Leftrightarrow\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1=\frac{3}{2}+3\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Leftrightarrow6.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)

\(\Rightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{9}{2}:6=\frac{3}{4}\)

Vậy \(P=\frac{3}{4}\)

a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

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

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)

Xét a+b+c=0

\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)

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

Xét a+b+c\(\ne0\)

\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)

\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{2c}{b}\cdot\frac{2a}{c}\cdot\frac{2b}{a}=8\)

Giải:
+) Xét a + b + c = 0

\(\Rightarrow-a=b+c\)

\(\Rightarrow-b=a+c\)

\(\Rightarrow-c=a+b\)

Ta có:

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

Lại có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=-1\)

+) Xét \(a+b+c\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Ta có:

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)

Vậy M = -1 hoặc M = 8

Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1

        c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1

=> A = 1+bc+b/bc+b+1 = 1

Tk mk nha

BÀI 1:

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

\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)        

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)       (thay   abc = 1)

\(=\frac{a+ab+1}{a+ab+1}=1\)

Chỉ cần bài 4 và bài 5 nữa thui ak

Ta có: \(Q+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{b+c}\right)+\left(1+\frac{c}{a+b}\right)\)

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

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

\(Q+3=2028\cdot\frac{1}{3}=676\)

=> Q = 676 - 3 = 673

xét a + b + c = 0 khi đó a + b = -c ; b + c = -a ; a + c = -b

Ta có : \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=-1\)

xét a + b + c \(\ne\)0 . thì \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow a+b=2c;b+c=2a\)\(\Rightarrow a-c=2\left(c-a\right)\)\(\Rightarrow a=c\)( loại vì a khác c )

Vậy A = -1

Ta có:

\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)

\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)

\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)

\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)

\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)

\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Vì a³+b³+c³=3abc và a+b+c khác 0

=>\(a^2+b^2+c^2-ab-bc-ca=0\)

\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Tổng 3 số không âm = 0 <=> chúng đều = 0

\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)

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

\(\)

Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)

b,

Ta có:  \(P=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=\frac{a+b}{ab}.\frac{b+c}{bc}.\frac{a+c}{ca}\)

Vì a=b=c (chứng minh tương tự câu a)

=>a+b=2b

 b+c=2c

 a+c=2a

Do đó \(P=\frac{2b}{ab}.\frac{2c}{bc}.\frac{2a}{ca}=\frac{\left(2.2.2\right).abc}{\left(abc\right)^2}=\frac{8}{abc}\left(đpcm\right)\)

Ta có S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)

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

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

\(=4000.\frac{1}{40}=100\)(a + b + c + d = 4000 ; \(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}=\frac{1}{40}\))

=> S = 100 - 4 = 96