Cho mình hỏi, phân thức cuối cùng của câu a phải là \(\frac{1}{c+2a+b}\)chứ

a)2a-3.-2+4=0=>2a+6+4=0=>2a=-10=>2=-5

b)3a-6-2.-1=2=>3a-6+2=2=>3a-6=0=>3a=6=>a=2

c)1-2.-3+-7-3a=-9=>1+6-7-3a=-9=>0-3a=-9=>3a=9=>a=3

b , 3a - b - 2c = 2 với b =  6 ; c = - 1

Ta có : 

3a - 6 - 2 . ( - 1 ) = 2

3a - 8 . ( - 1 ) = 2

3a + 8 = 2

3a       = 2 - 8

3a       = - 6

=> a = - 6 : 3

a = - 2

sai hết rùi

\(a,\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b\)

\(b,\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)

\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)

\(d,a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)\)

\(e,a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)\)

a) (a - b + c) - (a + c)

= a - b + c - a - c

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

= -b

b) (a + b) - (b - a) + c

= a + b - b + a + c

= 2a + (b - b) + c

= 2a + c

c) - (a + b - c) + (a - b - c)

= -a - b + c + a - b - c

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

= -2b

d) a(b + c) - a(b + d)

= ab + ac - ab - ad

= (ab - ab) + (ac - ad)

= ac - ad

= a(c - d)

e) a(b - c) + a(d + c)

= a(b - c + d + c)

= a[b - (c - c) + d]

= d(b + d)

a ) 

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

 a - b + c - a -c = - b

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

0 + 0 - b = -b ( đpcm )

Vân dụng bất đẳng thức \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(\Rightarrow\frac{1}{a+3b}+\frac{1}{b+2c+a}\ge\frac{4}{\left(a+3b\right)+\left(b+2c+a\right)}=\frac{2}{a+2b+c}\)

\(\Rightarrow\frac{1}{b+3c}+\frac{1}{c+2a+b}\ge\frac{4}{\left(b+3c\right)+\left(c+2b+a\right)}=\frac{2}{b+2c+a}\)

\(\Rightarrow\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{4}{\left(c+3a\right)+\left(a+2b+c\right)}=\frac{2}{c+2a+b}\)

Cộng tất cả các vế bất đẳng thức trên và rút gọn ta có bất đẳng thức \(\frac{1}{a+2b+c}+\frac{1}{b+2c+a}+\frac{1}{c+2a+b}\le\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\)

Đẳng thức xảy ra khi: \(\hept{\begin{cases}a+3b=b+2c+a\\b+3c=c+2a+b\Leftrightarrow a=b=c\\c+3a=a+2b+c\end{cases}}\)

Ta áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Áp dụng vào bài toán ta có : 

\(\frac{1}{a+3b}+\frac{1}{a+b+2c}\ge\frac{4}{a+3b+a+b+2c}=\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)

\(\frac{1}{b+3c}+\frac{1}{2a+b+c}\ge\frac{4}{b+3c+2a+b+c}=\frac{4}{2a+2b+4c}=\frac{2}{a+b+2c}\)

\(\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{4}{c+3a+a+2b+c}=\frac{4}{4a+2b+2c}=\frac{2}{2a+b+c}\)

Cộng vế theo vế của bất đẳng thức ta được 

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

=> ĐPCM

có a+b/b=k=>a+b=b.k=>b.k/b=k

     c+d/d=k=>c+d=d.k=>d.k/d=k

=>a+b/b=c+d/d

Bài 1 :

\(a,\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

Ta có : \(VT=\left(a-b\right)+\left(c-d\right)-\left(a-c\right)\)

                 \(=a-b+c-d-a+c\)

                 \(=-\left(b+d\right)=VP\)

\(\Rightarrow\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)

\(b,\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

Ta có : \(VT=\left(a-b\right)-\left(c-d\right)+\left(b+c\right)\)

                 \(=a-b-c+d+b+c\)

                 \(=a+d=VP\)

\(\Rightarrow\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)