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1/ (a – b + c) – (a + c) = -b
a-b+c-a-c=-b
-b=-b
2/ (a + b) – (b – a) + c = 2a + c
a+b-b+a+c=2a+c
2a+c=2a+c
3/ - (a + b – c) + (a – b – c) = -2b
-a-b+c+a-b-c=-2b
-(b.2)=-2b
-2b=-2b
4/ a(b + c) – a(b + d) = a(c – d)
ab+ac-ab+ad=a(c-d)
ac-ad=a(c-d)
a(c-d)=a(c-d)
5/ a(b – c) + a(d + c) = a(b + d)
ab-ac+ad+ac=a(b+d)
ab+ad=a(b+d)
a(b+d)=a(b+d)
6/ a.(b – c) – a.(b + d) = -a.( c + d)
ab-ac-ab=ad=-a(c+d)
-ac+ad=-a(c+d)
-a(c+d)=-a(c+d)
a, ( x + 5) + ( x – 9) = x + 2
x + 5 + x -9 = x+2
x+x-x = 2 -5 +9
x= 6
Bài 292: Tính
1 + 2 – 3 -4 + 5+ 6 -7 -8 +9 + ....+ 101 +102 -103 -104 +105.
Bài 293: Đơn giản các biểu thức sau:
a, A = ( a + b) + ( c –d) – ( a +c) – ( b-d)
= a+b+c -d -a-c -b +d
= 0
b, B = ( a –b) – ( c –d)- ( a + d) + ( b + c)
= a-b -c +d -a -d +b+c
= 0
a) (a - b - c) - (a - c)
= a - b - c - a + c
= (a - a) + (-b) + (-c + c)
= -b
b) (a - b) + (b - c) - (-c + a)
= a - b + b - c + c - a
= (a - a) + (-b + b) + (-c + c)
= 0
Ủng hộ mk nha !!! ^_^
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}\)\(=\frac{a+b+c+d}{a+b+c}\)
Do a + b + c + d khác 0 nên: b+c+d = a+c+d = a+b+d = a+b+c => a = b = c = d
\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)\(\left(a=b=c=d\right)\)
\(\Rightarrow A=1+1+1+1=4\)
\(=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2a+2b+2c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(A=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{2}{ac}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Linh không biết a + b + c = 0 để làm gì?
a) -(a-b+c)-(a+b+c)
= -a+b-c-a-b-c
= -(a-a)+(b-b)-(c-c)
= -0+0-0
= 0
b) a(b-c-d)-a(b+c-d)
= a.b-a.c-a.d-a.b-a.c+a.d
= (a.b-a.b)-(a.c-a.c)+(-a.d+a.d)
= 0-0+0
= 0
(Mình ko chắc là đúng nha)
CHÚC BẠN HỌC TỐT!
a, \(-\left(a-b+c\right)-\left(a+b+c\right)=-a+b-c-a-b-c=-2a-2c\)
b, \(a\left(b-c-d\right)-a\left(b+c-d\right)=a\left(b-c-d-b-c+d\right)=-2ac\)
c, \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)=ac+ad+bc+bd-ab-ac-bd-cd\)
\(=bc-cd=c\left(b-d\right)\)
a) A = ( a+ b – c + d ) – ( - a – b + c – d )
= a+ b – c + d + a + b - c + d
= 2a + 2b + 2d – 2c
b) B = ( a+ b – c ) + ( a – b ) – ( a – b - c )
= a+ b – c + a – b – a + b + c
= a + b