Ta có :\(\text{VT = A + B}\)

\(\text{= ( a + b + 5 ) + ( b – c – 9 )}\)

\(\text{= a + b + 5 + b – c – 9}\)

\(\text{= a + ( b + b ) – c + ( 5 – 9 )}\)

\(\text{= a + 2b – c – 4 (1)}\)

\(\text{VP = C – D}\)

\(\text{= ( b – c – 4 ) – ( -b – a )}\)

\(\text{= b – c – 4 + b + a}\)

\(\text{= ( b + b ) – c + a – 4}\)

\(\text{= 2b – c + a – 4}\)

\(\text{= a + 2b – c – 4 (2)}\)

\(\text{từ (1) và (2) suy ra}\)\(\text{ A + B = C – D ( đpcm ) }\)

b.(a+b)-(b-a)+c=2a+c

Xét VT: (a+b)-(b-a)+c = a + b - b + a + c = 2a+c

Mà VP = 2a+c

=> VT = VP 

c.-(a+b-c)+(a-b-c)=-2b

Xét VT: -(a+b-c)+(a-b-c) = -a - b + c + a - b - c = -2b

Mà VP = -2b

=> VT = VP

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

Xét VT: a(b+c)-a(b+d) = ab + ac - ab - ad =  ac - ad = a(c-d)

Mà VP = a(c-d)

=> VT = VP

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

Xét VT: a(b-c)+a(d+c)= ab -ac + ad + ac = ab + ad = a(b+d)

Mà VP = a(b+d)

=> VT = VP

1) (a-b+c)-(a+c)=a-b+c-a-c=-b (đpcm)

2) (a+b)-(b-a)+c=a+b-b+a+c=2a+c (đpcm)

3) -(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-2b (đpcm)

4) a(b+c) -a(b+d)=ab+ac-ab-ad=ac-ad=a(c-d) (đpcm)

5) a(b-c)+a(d+c)=ab-ac+ad+ac=ab+ad=a(b+d) (đpcm)

CHÚC BẠN HỌC TỐT NHÉ!

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

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

\(-b=-b\left(đpcm\right)\)

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

\(a+b-b+a+c=2a+c\)

\(2a+c=2a+c\left(đpcm\right)\)

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

\(-a-b+c+a-b-c=-2b\)

\(-2b=-2b\left(đpcm\right)\)

lm cx dễ thoi , bn lm tiếp nha ! 

1) (a – b + c) – (a + c) = -b

vế trái : (a – b + c) – (a + c)

=a - b + c - a + c

=a - a + c -c - b

=0 + 0 - b = - b

= - b vế phải

1. (a-b+c) -(a+c) = a-b+c-a-c = -b
2. (a+b) - (b-a) +c = a+b -b +a +c =2a+c
3. -(a+b-c)+(a-b-c) = -a-b+c a-b-c = -2b
4. a(b+c) -a(b+d) = a(b+c-b-d) = a( c-d)
5. a(b-c) +a(d+c) = a(b-c+d+c) = a(b+d)

1.= a-b+c-a-c= (a-a)-b+(c-c)=0-b+0=-b

2.=a+b-b+a+c=a+a+b-b+c=2a+c

3.=-a-b+c+a-b-c=-a+a-(b+b)+c-c=-2b

4.=ab+ac-ab-ad=ac-ad=a(c-d)

5.=ab-ac+ad+ac=(-ac+ac)+ab+ad=ab+ad=a(b+d)

tk mik nha, chúc bn học tốt
 

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

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

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

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

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

1/ (a-b+c) - (a+c) = a-b+c-a-c = -b 
2/ (a+b) - (b-a) + c = a + b - b + a + c = 2a + c
3/ - (a+b-c) + (a-b-c) = -a - b + c + a - b - c = - 2b
4/ a(b+c) - a(b+d) = a(b+c-b-d) = a(c-d)
5/ a(b-c) + a(d+c) = a(b-c+d+c) = a(b+d)
 

1/   (a-b+c) - (a+c) = a-b+c -a-c

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

                              = 0+0-b

                              = -b

2/   (a+b) - (b-a) +c= a+b -b+a +c

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

                              = 2a+c

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

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

                                 = 0 +(-2b)+ 0

                                 = -2b

4/   a (b+c)- a(b+d) = ab+ac - ab-ad

                               = (ab-ab) +(ac-ad)

                               = 0+ a.(c-d)

                               = a.(c-d)

5/ a(b-c) + a(d+c) = ab-ac + ad +ac

                             = (-ac+ac) + (ab+ad)

                             = 0+ a.(b+d)

                             = a.(b+d)

1/ Ta có:

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

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

                              =0 + 0 + (-b)

                              =-b

2/ Ta có:

(a+b)-(b-a)+c=a+b-b+a+c=(a+a)+(b-b)+c=2a+0+c=2a+c

3/ Ta có:

-(a+b-c)+(a-b-c)=-a-b+c+a-b-c=(-a+a)+(-b-b)+(c-c)=0+(-2b)+0=-2b

4/ Ta có:

a(b+c)-a(b+d)=ab+ac-ab-ad=(ab-ab)+(ac-ad)=0+(ac-ad)=ac-ad=a(c-d)

5/ Ta có:

a(b-c)+a(d+c)=ab-ac+ad+ac=(ab+ad)+(-ac+ac)=a(b+d)+0=a(b+d)

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)