chứng tỏ
1.(a-b+c)-(a+c)=-b
2.(a+b)-(b-a)+c=2a+c
3.-(a+b-c)+(a-b-c)=-2b
4.a(b+a)-a(b+d)=a(c-d)
5.a(b-c)+(d+c)=a.(b+d)
giúp mik vs
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(a-b+c)-(a+c)
= a-b+c-a-c
=(a-a)+(c-c)-b
=-b
2.( a + b ) - ( b - a ) + c
= a + b - b + a + c
=( a + a ) + ( b -b ) + c
= 2a + 0 + c
= 2a + c
mấy câu sau bn tự lm nha
(a - b + c) - (a + c) = a - b + c - a - c = -b (đpcm)
(a + b) - (b - a) + c = a + b - b + a + c = 2a + c (đpcm)
-(a + b - c) + (a - b - c) = -a - b + c + a - b - c = -2b (đpcm)
a.(b + c) - a.(b + d) = a.(b + c - b - d) = a.(c - d) (đpcm)
a.(b - c) + a.(d + c) = a.(b - c + d + c) = a.(b + d) (đpcm)
\(\text{ (a-b+c)-(a+c)}=a-b+c-a-c=\left(a-a\right)-b+\left(c-c\right)=-b\)
\(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab+ad=ac+ad=a\left(c+d\right)\)
\(a\left(b-c\right)+a\left(d+c\right)=a\left(b-c+d+c\right)=a\left(b+d\right)\)
Ta có \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
=> \(1+\frac{a+b+c+d}{a}=1+\frac{a+b+c+d}{b}=1+\frac{a+b+c+d}{c}=1+\frac{a+b+c+d}{d}\)
=> \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Khi a + b + c + d => a + b = -(c + d) ;
b + c = -(a + d) ;
c + d = -(a + b)
d + a = -(b + c)
Khi đó \(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)
= (-1) + (-1) + (-1) + (-1) = -4
Khi a + b + c + d \(\ne0\)
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = \(\frac{2a}{2a}+\frac{2b}{2b}+\frac{2c}{2c}+\frac{2d}{2d}=2+2+2+2=8\)
Vậy khi a + b + c + d = 0 thì M = -4
khi a + b + c + d \(\ne\)0 thì M = 8
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)
a)\(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\)
\(\Leftrightarrow ac-bc+ad-bd=ac-ad+bc-bd\)
\(\text{Thay }ad=bc\text{ vào}\Rightarrow ac-ad+ad-bd=ac-ad+ad-bd\)
\(\text{Đây là đẳng thức đúng }\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\text{ là đúng }\)
b)\(\text{Tương tự*}\)
a) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{b}{a+b}=\frac{d}{c+d}\)
\(\Leftrightarrow\frac{-2b}{a+b}+1=\frac{-2d}{c+d}+1\Leftrightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{4a}{b}-5=\frac{4c}{d}-5\Leftrightarrow\frac{4a-5b}{b}=\frac{4c-5d}{d}\Leftrightarrow\frac{b}{4a-5b}=\frac{d}{4c-5d}\)
\(\Leftrightarrow\frac{11b}{4a-5b}+1=\frac{11d}{4c-5d}+1\Leftrightarrow\frac{4a+6b}{4a-5b}=\frac{4c+6d}{4c-5d}\Leftrightarrow\frac{2a+3b}{4a-5b}=\frac{2c+3d}{4c-5d}\)
\(\Leftrightarrow\frac{2a+3b}{2c+3d}=\frac{4a-5b}{4c-5d}\)
1,( a + b ) - ( b - a) +c
= a + b - b + a + c
= ( a + a ) + ( b - b ) + c
= 2a + c
2. - ( a + b - c) + ( a - b - c )
= -a -b +c + a - b - c
= ( -a + a ) - ( b + b ) + ( c - c )
= -2b
mấy câu sau bn tự giải nhá. MỆT
3,-(a+b-c)+(a-b-c)
=-a-b+c+a-b-c
=(-a+a)-b-b+(c-c)
=0-b-b+0
=-b-b
=-2b(đpcm)
4,a(b+c)-a(b+d)
=ab+ac-ab+ad
=(ab-ab)+ac+ad
=0+ac+ad
=ac+ad
=a(c+d)(đpcm)
5,a(b-c)+a(d+c)
=ab-ac+ad+ac
=(-ac+ac)+ab+ad
=0+ab+ad
=ab+ad
=a(b+d)(đpcm)
k cho mình vs
1. ( a - b + c ) - ( a + c ) = - b
Ta có : VT = ( a - b + c ) - ( a + c )
= a - b + c - a - c
= - b = VP
=> ( a - b + c ) - ( a + c ) = - b ( đpcm )
2) ( a + b ) - ( b - a ) + c = 2a + c
Ta có : VT = ( a + b ) - ( b - a ) + c
= a + b - b + a + c
= 2a + c = VP
=> ( a + b ) - ( b - a ) + c = 2a + c ( đpcm )
3) - ( a + b - c ) + ( a - b - c ) = - 2b
Ta có : VT = - ( a + b - c ) + ( a - b - c )
= - a - b + c + a - b - c
= - 2b = VP
=> - ( a + b - c ) + ( a - b - c ) = - 2b ( đpcm )
4) a( b + c ) - a ( b + d ) = a ( c - d )
Ta có : VT = a ( b + c ) - a ( b + d )
= ab + ac - ab - ad
= ac - ad
= a ( c - d ) = VP
=> a( b + c ) - a( b + d ) = a( c - d ) ( đpcm )
5) a( b - c ) + a( d + c ) = a( b + d )
Ta có : VT = a( b - c ) + a ( d + c )
= a ( b - c + d + c )
= a( b + d ) = VP
=> a ( b - c ) + a ( d + c ) = a ( b + d ) ( đpcm )
VT là vế trái
VP là vế phải .