chứng minh:
(a-b)+(c-d)=(a+c)-(b+d)
(a-c)-(c-d)=(a+d)-(a+d)-(b+c)
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=a/c chứng minh rằng a/c=a+b/c+d
b. a/b=c/d chứng minh rằng a/c=a-b/c-d
c. a/b=c/d chứng minh rằng a+b/a-b=c+d/c-d
Giúp em nó😊😊
Cho a+b+c+d=0
a) Chứng minh a^3+b^3+c^3+d^3=3(ab-cd)(c+d)
b)Chứng minh (a+b+c+)^3=a^3 + b^3 + c^3+3(a+b)(b+c)(c+a)
c)Cho c-a=b+d. Chứng Minh a^3+b^3-c^3+d^3=3(d-c)(ab+cd)
a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)
Cho a/b=c/d. Chứng minh:
1) (a+c)b=(b+d)a
2) (b+d)c=(a+c)d
3)(a+b)(c-d)=(a-b)(c+d)
Do a/b=c/d ⇔ ad=bc
1) Ta có: (a+c)b=ab+bc
(b+d)a=ab+ad
Do bc=ad nên ab+ad=ab+bc
Suy ra (a+c)b=(b+d)a (đpcm)
2) Ta có: (b+d)c=bc+dc
(a+c)d=ad+cd
Do bc=ad nên bc+dc=ad+cd
Suy ra (b+d)c=(b+d)c (đpcm)
3)Ta có:(a+b)(c-d)=ac-ad+bc-bd=(ac-bd)-(ad-bc)
(a-b)(c+d)=ac+ad-bc-bd=(ac-bd)+(ad-bc)
Do ad=bc ⇔ ad-bc=0 nên (ac-bd)-(ad-bc)=(ac-bd)+(ad-bc)
⇔(a+b)(c-d)= (a-b)(c+d) (đpcm)
Từ tỉ lệ thức a/b = c/d, hãy chứng minh: a) a + b/b = c + d/d ; b) a - b/b = c - d/d ; c) a + b/a = c + d/c ; d) a - b/a = c - d/c
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Rightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm1\right).\)
b) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)
\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm2\right).\)
c) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)
\(\Rightarrow\frac{b}{a}+\frac{a}{a}=\frac{d}{c}+\frac{c}{c}\)
\(\frac{b+a}{a}=\frac{d+c}{c}\left(đpcm3\right).\)
d) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)
\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}\)
\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\left(đpcm4\right).\)
Chúc bạn học tốt!
Chứng minh rằng : (a+b+c-d)(a-b-c-d) = (a+b-c+d)(a-b+c+d) thì (a+b)/(a-b) = (c-d)/(c+d)
Chứng minh: c,a-b/b=c-d/d d,a+c/b+d=a-c/b-d
c: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a-b}{b}=\dfrac{bk-b}{b}=k-1\)
\(\dfrac{c-d}{d}=\dfrac{dk-d}{d}=k-1\)
Do đó: \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
CHỨNG MINH ĐẲNG THỨC
A) a.(b+c) - a.(b+d)= a.(c-d)
B) a.(b-c) + a.(d-c)= a.(b+d)
C) a.(b-c) - a.(b+d)= -a.(c+d)
D) (a+b).(c+d)-(a+b).(b+c)= (a-c).(d-b)
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,
Ta có: a.(b+c) - a.(b+d)
= ab+ac-ab-ad
= (ab-ab)+(ac-ad)
= ac-ad
= a.(c-d)
b, Phần này phải là a.(b-c) + a.(d+c) mới đúng nha
Ta có: a.(b-c) + a.(d-+c)
= ab-ac+ad+ac
= (ac-ac)+(ab+ad)
= ab+ad
= a.(b+d)
c,
Ta có: a.(b-c) - a.(b+d)
= ab-ac-ab-ad
= (ab-ab)-(ac-ad)
= -ac + ad
= -a.(c+d)
CHỨNG MINH ĐẲNG THỨC
A) a.(b+c) - a.(b+d)= a.(c-d)
B) a.(b-c) + a.(d-c)= a.(b+d)
C) a.(b-c) - a.(b+d)= -a.(c+d)
D) (a+b).(c+d)-(a+b).(b+c)= (a-c).(d-b)
A) a.(b + c) - a.(b + d) = a.b + a.c - a.b - a.d B) a.(b - c) + a.(d - c) = a.b - a.c + a.d - a.c
= (a.b - a.b) + (a.c - a.d) = (a.b + a.d) - (a.c - a.c)
= a.c - a.d = a.(b + d) - a.c + a.c
= a.(c - d) = a.(b + d)
C) a.(b - c) - a.(b + d) = a.b - a.c - a.b + a.d
= (a.b - a.b) - (a.c + a.d)
= 0 - a.(c + d)
= -a.(c + d)
chứng minh rằng : (a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d) thì a/c=b/d
\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left[\left(a+d\right)+\left(b+c\right)\right]\left[\left(a+d\right)-\left(b+c\right)\right]\)
\(=-\left(b+c\right)^2+\left(a+d\right)^2\) ( 1 )
\(\left(a+b-c-d\right)\left(a-b+c-d\right)=\left(b-c\right)^2-\left(a-d\right)^2\) ( 2 )
Từ ( 1 ) và ( 2 ) suy ra
\(b^2+2bc+c^2-a^2-2ad-d^2=a^2-2ad+d^2-b^2+2bc-c^2\)
\(4ad=4ac\Rightarrow ad=bc\)
\(\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}\)( đpcm )