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Bài 1 :
Ta có : P = a.{ ( a - 3 ) - [(a+3) - [ ( a + 2 ) - (a - 2 )]}
= a . { ( a - 3 ) - [ ( a + 3 ) - ( -a - 2 )]}
= a . ( a - 3 -a - 3 - a + 2 )
= a . ( - a - 8 ) = -8a -a2
: Q = [a +( a + 3 ) ] - [ ( a + 2 ) - ( a - 2 ) ]
= a + a + 3 - a - 2 - a - 2
= -1
Ta thấy -1> -8a - a2 => Q > P
Bài 2 :
Ta có : a - ( b - c ) = ( a - b ) + c = ( a + c ) - b
<=> a - b + c = a - b + c = a + c - b
do a = a ; b = b ; c = c => 3 vế bằng nhau (đpcm)
Bài 3:
a) ( a - b ) + ( c - d ) = ( a + c ) - ( b + d )
<=> a - b + c - d = a + c - b - d
<=> a - a + c - c - b + b - d + d = 0
<=> 0 = 0 => VP = VT ( đpcm)
b) a - b - ( c- d ) = ( a + d ) - ( b + c )
<=> a - b - c + d = a + d - b -c
<=> a - a - b + b - c + c + d -d = 0
<=> 0 =0 => VP = VT ( đpcm )
1, Chứng minh đẳng thức :
a) (a - b + c) - (a + c) = -b
(a - b + c) - (a + c)
=a-b+c-a-c
=(a-a)+(c-c)-b
=0+0-b
=-b
b) (a + b) - (b - a) + c = 2a + c
(a + b) - (b - a) + c
=a+b-b+a+c
=(a+a)+(b-b)+c
=2a+0+c
=2a+c
c) -( a + b - c) + (a- b- c) = -2b
-( a + b - c) + (a- b- c)
=-a-b+c+a-b-c
=[a+(-a)]+[c+(-c)]-b-b
=0+0-(b+b)
=-2b
d) a( b+c) - a (b +d) =a( c-d )
a( b+c) - a (b +d)
=ab+ac-(ab+ad)
=(ab-ab)+ac-ad
=0+ac-ad
=a(c-d)
e) a (b - c) + a( d+ c) = a( b+d)
a (b - c) + a( d+ c)
=ab-ac+ad+ac
=(ac+(-ac))+ad+ab
=0+ad+ab
=a(d+b)
1
a) \( (a - b + c) - (a + c) \)
\(=\left(a+c-b\right)-\left(a+c\right)\)
\(=\left[\left(a-c\right)-\left(a-c\right)\right]-b\)
\(=0-b\)
\(=-b\)
b) \( (a + b) - (b - a) + c \)
\(=a+b-b+a+c\)
\(=\left(a+a\right)+\left(b-b\right)+c\)
\(=\left(a+a\right)-0+c\)
\(=a+a+c\)
\(=2a+c\)
2
\(P=a+ [( a - 3 ) - (-a - 2)]\)
\(P=a+a-3+a+2\)
\(P=a+a+a-3+2\)
\(P=3a-3+2\)
\(P=0+2\)
\(P=2\)
\(Q=[a + (a +3)] - [( a + 2) - ( a - 2)]\)
\(Q=a+a+3-a-2-a+2\)
\(Q=a+a+3-a+\left(-2-a+2\right)\)
\(Q=2a+3-a+a\)
\(Q=2a+3-2a\)
\(Q=3\)
Vì \(P=2;Q=3\Rightarrow P< Q\)
1)-(a+b-c)+(a-b-c)=a-b+c+a-b-c=-2b
2)a(b+c)-a(b+d)=a(b+c-b-d)=a(c-d)
3)a(b-c)+a(d+c)=a(b-c+d+c)=a(b+d)
chúc hok tốt :))))))
ak ở câu 1 sửa lại chút
1)-(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-2b
a, =-25.21.4.(-3).(-1)
=-25.4.21.4
=-100.21.4
=-2100.4
=-8400
b, =-125.67.(-8).1
=-125.(-8).67
=1000.67
=67000
c, =35.18-35.28
=35.(18-28)
=35.(-10)
=-350
d, =24.11-16.24+16.5
=24.(11-16)+16.5
=24.(-5)+16.5
=5.(-24+16)
=5.(-8)
=-40
e, =29.6-19.29+19.13
=29.(6-19)+19.13
=29.(-13)+19.13
=13.(-29+19)
=13.(-10)
=-130
Bài 17 :
1) ab + ac = a ( b + c )
2) ab - ac + ad = a ( b - c + d )
3) ax - bx - cx + dx = x ( a- b - c + d )
4) a(b + c) – d(b + c) = ( b + c ) ( a - d )
5) ac – ad + bc – bd = a( c - d ) + b ( c - d ) = ( c- d ) ( a + b )
6) ax + by + bx + ay = a( x+ y ) + b ( x + y ) = ( x + y ) (a +b )
Bài 18:
1/ (a – b + c) – (a + c) = a - b + c - a - c = -b
2/ (a + b) – (b – a) + c = a + b - b + a + c = 2a + 2
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=-b
2,a+b-b-a+c=2a+c
3,-a-b+c+a-b-c=-2b
4,ab+ac-ab-ad=ac-ad=a(c-d)
5,ab-ac+ad+ac=ab+ad=a(b+d)
lớp 6 có cm đẳng thức hử?
a,(a-b-c)-(a+c)=-b
suy ra:a-b-c-a-c=-b
(a-a)-(c-c)-b=-b
0-b=-b