Cho $\frac{a}{b+c+d}$ = $\frac{b}{c+d+a}$ = $\frac{c}{d+a+b}$ = $\frac{d}{a+b+c}$
Tính giá trị biểu thức: S= $\frac{a+b}{c+d}$ + $\frac{b+c}{d+a}$ + $\frac{c+d}{a+b}$+ $\frac{d+a}{b+c}$
Cho $\frac{a}{b+c+d}$ = $\frac{b}{c+d+a}$ = $\frac{c}{d+a+b}$ = $\frac{d}{a+b+c}$
Tính giá trị biểu thức: S= $\frac{a+b}{c+d}$ + $\frac{b+c}{d+a}$ + $\frac{c+d}{a+b}$+ $\frac{d+a}{b+c}$
Đáp án:
`S=4\or\S=-4`
Giải thích các bước giải:
`a+b+c+d=0`
`->a+b=-(c+d)\and\b+c=-(d+a)\and\c+d=-(a+b)\and\d+a=-(b+c)`
`->S=(-(c+d))/(c+d)+(-(d+a))/(d+a)+(-(a+b))/(a+b)+(-(b+c))/(b+c)`
`->S=-1-1-1-1`
`->S=-4`
`a+b+c+d ne 0`
Áp dụng tính chất dãy tỉ số bằng nhau ta có
`a/(b+c+d)+b/(c+d+a)+c/(d+a+b)+d/(a+b+c)=(a+b+c+d)/(3a+3b+3c+3d)=1/3`
`->3a=b+c+d`
`->4a=a+b+c+d`
Hoàn toàn tương tự:
`4b=a+b+c+d`
`4c=a+b+c+d`
`4d=a+b+c+d`
`->4a=4b=4c=4d->a=b=c=d`
`->S=(2a)/(2a)+(2b)/(2b)+(2c)/(2c)+(2d)/(2d)`
`->S=1+1+1+1`
`->S=4`
Vậy `S=4\or\S=-4`
`cancel{nocopy//2072007}`
Đáp án:
$+)\quad M = -4\quad khi \quad a+b+c+d = 0$
$+)\quad M = 4\,\,\,\quad khi\quad a+b+c+d\ne 0$
Giải thích các bước giải:
$+)\quad a+b+c+d = 0$
$\to \begin{cases}a+b = – (c+d)\\b+c= -(d+a)\\c+d = -(a+b)\\d+a = -(b+c)\end{cases}$
$\to M = \dfrac{-(c+d)}{c+d}+\dfrac{-(d+a)}{d+a}+\dfrac{-(a+b)}{a+b}+\dfrac{-(b+c)}{b+c}$
$\to M = -1-1-1-1$
$\to M = -4$
$+)\quad a+b+c+d \ne 0$
Ta có:
$\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}$
$\to \dfrac{a}{b+c+d}+1=\dfrac{b}{c+d+a}+1=\dfrac{c}{d+a+b}+1=\dfrac{d}{a+b+c}+1$
$\to \dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{c+d+a}=\dfrac{a+b+c+d}{d+a+b}=\dfrac{a+b+c+d}{a+b+c}$
$\to \dfrac{1}{b+c+d}=\dfrac{1}{c+d+a}=\dfrac{1}{d+a+b}=\dfrac{1}{a+b+c}$
$\to b+c+d = c +d + a = d+a+b = a+b+c$
$\to a = b = c = d$
Do đó:
$M =\dfrac{a+a}{a+a} +\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}$
$\to M = 1+1+1+1$
$\to M = 4$