bài 1:cho A=a+b-5 B=-b-c+1 C=b-c-4 D=b-a
Chứng minh: A+B=C+D
bài 2:chứng minh
a/ (a-b+c)-(a+c)=-b
b/(a+b)-(b-a)+c=2a+c
c/(a+b-c)+(a-b-c)=-2b
d/a(b+c)-a(b-d)=a(c-d)
giải giúp mình với!
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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) = 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) 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