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\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}\)\(=\frac{a+b+c+d}{a+b+c}\)
Do a + b + c + d khác 0 nên: b+c+d = a+c+d = a+b+d = a+b+c => a = b = c = d
\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)\(\left(a=b=c=d\right)\)
\(\Rightarrow A=1+1+1+1=4\)
ta nhân lần lượt a,b,c,d vào biểu thức ban đầu , được
\(\hept{\begin{cases}\frac{a^2}{b+c+d}+\frac{ba}{a+c+d}+\frac{ac}{a+b+d}+\frac{ad}{a+b+c}=a\left(1\right)\\\frac{ab}{b+c+d}+\frac{b^2}{a+c+d}+\frac{cb}{a+b+d}+\frac{db}{a+b+c}=b\left(2\right)\end{cases}}\)
\(\hept{\begin{cases}\frac{ac}{b+c+d}+\frac{bc}{c+a+d}+\frac{c^2}{a+b+d}+\frac{dc}{a+b+c}=c\left(3\right)\\\frac{ad}{b+c+d}+\frac{bd}{a+c+d}+\frac{cd}{a+b+d}+\frac{d^2}{a+b+c}=d\left(4\right)\end{cases}}\)
Lấy (1)+(2)+(3)+(4) ta có :
\(\left(\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}\right)+\frac{ab+bc+bd}{c+d+a}+\frac{ac+bc+cd}{d+a+b}\)
\(+\frac{ad+bd+cd}{a+b+c}+\frac{ab+ac+ad}{b+c+d}=a+b+c+d\)
\(< =>A+\frac{b\left(c+d+a\right)}{c+d+a}+\frac{d\left(a+b+c\right)}{a+b+c}+\frac{c\left(b+d+a\right)}{b+d+a}+\frac{a\left(c+b+d\right)}{c+b+d}=a+b+c+d\)
\(< =>A+a+b+c+d=a+b+c+d=>A=0\)
Vậy \(A=\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}=0\)
a, -( -a + c - d) - ( c - d + d) = a - c + d - c + d - d = a + d
b, - ( a+b-c+d) + (a-b-c-d) = -a -b+c-d + a-b-c-d = -2b + (-2c)= -2(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 + c – d ) – ( c – a + d )
= a - c - d - c + a + d
= (a + a) + (-c - c) + (-d + d)
= 2a - 2c
b) – ( a + b - c + d ) + ( a – b – c –d )
= - a - b + c - d + a - b - c - d
= (-a + a) + (-b - b) + (c - c) + (-d - d)
= -2b - 2d
a) - ( - a + c - d) - ( c - a + d )
= a - c + d - c + a - d
= 2a
b) - ( a+ b - c + d ) + ( a -b -c -d )
= - a-b+c-d+a-b-c-d
=-2d -2b
c) a(b-c-d) - a(b+c-d)
= a(b-c-d-b-c+d)
= ab-ac-ad-ab-ac+ad
= -2ab-2ac
d) (a+b)(c+d)-(a+d)(b+c)
= ac+ad+bc+bd - (ab+ac+bd+cd)
= ac+ad+bc+bd-ab-ac-bd-cd
=ad+bc-ab-cd
minh châu oi