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A) \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt,c=dt\)
\(\frac{a}{a+b}=\frac{bt}{bt+b}=\frac{t}{t+1},\frac{c}{c+d}=\frac{dt}{dt+d}=\frac{t}{t+1}\)
suy ra đpcm.
\(\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b}{d},\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b}{d}\)
suy ra đpcm.
B) \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-\left(a+c\right)}{\left(b+3d\right)-\left(b+d\right)}=\frac{2c}{2d}=\frac{c}{d}\)
\(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-3\left(a+c\right)}{\left(b+3d\right)-3\left(b+d\right)}=\frac{-2a}{-2b}=\frac{a}{b}\)
suy ra đpcm.
A) a.(b+c) - a.(b+d)= a.(c-d)
=> ab+ac -ab-ad=ac-ad
=>ac-ad=ac-ad(đpcm)
các câu kia bạn lm tương tự
bn vào câu hỏi tương tự và tìm câu hỏi của trần thị mỹ trang tham khảo
a. (a-b)+(c-d)=(a+c)-(b+d)
Ta có: VP=(a+c)-(b+d)=a+c-b-d=(a-b)+(c-d)=VT
=> VT=VP (đpcm)
b. Ta có: VT=a(b+c)-b(a-c)=ab+ac-ab+bc=ac+bc=c(a+b)=VP
=> VT=VP (đpcm)
c. Ta có: VT=(a+b)(c+d)-(a+d)(b+c)=ac+ad+bc+bd-ab-ac-bd-cd=ad+bc-ab-cd
VP=(a-c)(d-b)=ad-ab-cd+bc=ad+bc-ab-cd=VT
=> VT=VP (đpcm)
\(\text{ (a-b+c)-(a+c)}=a-b+c-a-c=\left(a-a\right)-b+\left(c-c\right)=-b\)
\(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab+ad=ac+ad=a\left(c+d\right)\)
\(a\left(b-c\right)+a\left(d+c\right)=a\left(b-c+d+c\right)=a\left(b+d\right)\)
1) a( b+c) - b(a-c) = ( a+b) c
VT = a( b+c) - b(a-c)
= ab + ac - ab + bc
= ac + bc
= c(a + b) (=VP)
2)a (b - c)- a (b+d)= - a (c+d)
VT= a (b - c)- a (b+d)
= ab - ac - ab - ad
= -ac - ad
= -a(c + d) (=VP)
1, a(b+c)-b(a-c)=(a+b)c
\(ab+ac-ba+bc=\left(a+b\right)c\)
\(a.\left(b-b\right)+\left(a+b\right).c=\left(a+b\right)c\)
\(a.0+\left(a+b\right)c=\left(a+b\right)c\)
\(\left(a+b\right)c=\left(a+b\right)c\)
\(\Rightarrowđpcm\)
2, a(b-c)-a(b+d)=-a(c+d)
\(ab-ac-ab-ad=a.\left(c+d\right)\)
\(a.\left(b-c-b-d\right)=a\left(-c-d\right)\)
\(a.\left(-c-d\right)=a.\left(-c-d\right)\)
\(\Rightarrowđpcm\)
3, (a+b)(c+d)-(a+d)(b+c)=(a-c)(d-b)
=ac+ad+bc+bd-ab-ac-bd-dc
=ad-ab+bc-dc
=(ad-ab)+(bc-dc)
=a(d-b)+c(b-d)
=a(d-b)-c(d-b)
=(a-c)(d-b) =VP.
\(\Rightarrowđpcm\)
học tốt
1,a.(b+c)-b.(a-c)
=a.b+a.c-(b.a-b.c)
=a.b+a.c-b.a+b.c
=(a.b-b.a)+(a.c+b.c)
=0+c.(a+b)=c.(a+b)
2)a.(b-c)-a.(b+d)
=a.b-a.c-(a.b+a.d)
=a.b-a.c-a.b-a.d
=(a.b-a.b)-a.c-a.d
=0-a.c-a.d
=-a.c-a.d
=-a.c+(-a.d)
=-a.(c+d)
3)(a+b).(c+d)-(a+d).(b+c)
=a.c+a.d+a.c+a.d-(a.b+a.c+d.b+d.c)
=a.c+a.d+a.c+b.d-a.b-a.c-d.b-d.c
=(a.c-a.c)+(b.d-d.b)+a.d+a.c-a.b-d.c
=0+0+(a-c).(d-b)
=(a-c).(d-b)
a) \(VT=a\left(b-c\right)-a\left(b+d\right)=a\left(b-c-b-d\right)=-a\left(c+d\right)=VP\)
b) \(VT=\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)=ac+ad+bc+bd-ab-ac-bd-dc\)
\(=ad+bc-ab-dc=a\left(d-b\right)-c\left(d-b\right)=\left(d-b\right)\left(a-c\right)=VP\)
p/s: chúc bạn học tốt
a) \(a\left(b-c\right)-a\left(b+d\right)=ab-ac-ab-ad=-ac-ad=-a\left(c+d\right)\)
=> ĐPCM
b) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\)
= a.(c+d)+b(c+d)-[a(b+c)+d(b+c)]
= ac+ad+bc+bd-(ab+ac+bd+cd)
= ac+ad+bc+bd-ab-ac-bd-cd
= ad+bc-ab-cd
= a(d-b)-c(d-b)
= (a-c)(d-b)
=> ĐPCM
a)(a-b+c)-(a+c)
=a-b+c-a-c
=a-a - b + c-c
=-b
b)(a+b)-(b-a)+c
=a+b-b+a+c
=a+a+c
=2a+c
c)-(a+b-c) +(a-b-c)
= -a -b+c+a-b-c
=-a+a-b-b+c-c
=-2b
d) a(b+c) - a(b+d)
= ab + ac - ab - ad
= ab-ab+ac-ad
=a(c-d)
e) a(b-c) + a(d+c)
= ab -ac + ad + ac
= ab +ad -ac + ac
=a(b+d)
a(b+c) - a(b+d)=a(c-d)
VP= a(b+c) - a(b+d)
= ab+ac-ab-ad
= ac-ad
=a(c-d)
Suy ra VP=VT
a(b-c)+a(d+c)=a(b+d)
VP= a(b-c)+a(d+c)
= ab-ac+ad+ac
= ab+ad
= a(b+d)
Suy ra VP=VT