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Ta có S = \(\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\)
=> S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{d+a+b}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)
= \(\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{c+d+a}+\frac{a+b+c+d}{d+a+b}+\frac{a+b+c+d}{a+b+c}\)
\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right)\)
\(=4000.\frac{1}{40}=100\)
=> S = 100 - 4 = 96
Em có cách khác!
\(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)
\(\Rightarrow\frac{a+b+c+d}{a+b+c}+\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{c+d+a}\)
\(+\frac{a+b+c+d}{d+a+b}=50\)
\(\Rightarrow\frac{d}{a+b+c}+1+\frac{a}{b+c+d}+1+\frac{b}{c+d+a}+1\)
\(+\frac{c}{d+a+b}+1=50\)
\(\Rightarrow\frac{d}{a+b+c}+\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}=46\)
Đề: \(a+b+c+d=2000\)
\(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)
Tính:
\(S=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\)
Giải:
Có: \(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)
=> \(\frac{1}{2000-d}+\frac{1}{2000-a}+\frac{1}{2000-b}+\frac{1}{2000-c}=\frac{1}{40}\)
<=> \(\frac{2000}{2000-d}+\frac{2000}{2000-a}+\frac{2000}{2000-b}+\frac{2000}{2000-c}=\frac{2000}{40}\)
<=> \(1+\frac{d}{2000-d}+1+\frac{a}{2000-a}+1+\frac{b}{2000-b}+1+\frac{c}{2000-c}=50\)
<=> \(\frac{d}{a+b+c}+\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}=46\)
=> \(S=46\)
Lời giải:
Áp dụng TCDTSBN:
$\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}$
$=\frac{a+b+c-d+b+c+d-a+c+d+a-b+d+a+b-c}{d+a+b+c}$
$=\frac{2(a+b+c+d)}{a+b+c+d}=2$
$\Rightarrow a+b+c-d=2d; b+c+d-a=2a; c+d+a-b=2b; d+a+b-c=2c$
$\Rightarrow a+b+c=3d; b+c+d=3a; c+d+a=3b; d+a+b=3c$
Khi đó:
\(P=\frac{a+b+c}{a}.\frac{b+c+d}{b}.\frac{c+d+a}{c}.\frac{a+b+d}{d}\\ =\frac{3d}{a}.\frac{3a}{b}.\frac{3b}{c}.\frac{3c}{d}=81\)
\(\frac{a+b+c-2d}{a}=\frac{b+d+a-2c}{b}=\frac{b+d+c-2a}{c}=\frac{a+c+d-2b}{d}\)
\(=\frac{\left(a+b+c-2d\right)+\left(b+d+a-2c\right)+\left(b+d+c-2a\right)+\left(a+c+d-2b\right)}{a+b+c+d}\)
\(=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Leftrightarrow a=b=c=d\).
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{d}\right)\left(1+\frac{d}{a}\right)=2^4=16\)
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)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+b}{a}=\dfrac{bk+b}{bk}=\dfrac{k+1}{k}\)
\(\dfrac{c+d}{c}=\dfrac{dk+d}{dk}=\dfrac{k+1}{k}\)
Do đó: \(\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
b: \(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{k-1}{k}\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{k-1}{k}\)
Do đó: \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
c: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
1) a.ended
2) c.pavement
3) c.obey
4) c. character
5) d. who
hok tốt
1.
a. ended b. loved c. claned
2.
a. way b. station c. pavement
3.
a. seatbelt b. enter c. obey d. present
4.
a. chair b. watch c. charater
5.
a. where b. whe c. what d. who