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1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
a)\(\left(x-\frac{1}{2}\right)^{2016},\left|\frac{3}{4}-y\right|\ge0\)
\(\left(x-\frac{1}{2}\right)^{2016}+\left|\frac{3}{4}-y\right|=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{2}\right)^{2016}=0\\\left|\frac{3}{4}-y\right|=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{4}-y=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{4}\end{cases}}\)
b)\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}\)
\(\Rightarrow\frac{b+c}{a}-\frac{a+c}{b}-\frac{a+b}{c}=0\)
Trường hợp 1: a+b+c \(\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left[a+b+c\right]}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{b+c}=\frac{1}{2}\Leftrightarrow\frac{b+c}{a}=2\\\frac{b}{a+c}=\frac{1}{2}\Leftrightarrow\frac{a+c}{b}=2\\\frac{c}{a+b}=\frac{1}{2}\Leftrightarrow\frac{a+b}{c}=2\end{cases}\Rightarrow}\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2+2+2=6\)
Trường hợp 2: a + b + c = 0
\(a+b+c=0\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}}\)
\(\Rightarrow\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=-\frac{a}{a}+-\frac{b}{b}+-\frac{c}{c}=-1+\left[-1\right]+\left[-1\right]=-3\)
Ta có :
\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{a+c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2c=a+b\\2b=a+c\end{cases}\Rightarrow\hept{\begin{cases}3a=a+b+c\\3c=a+b+c\\3b=a+b+c\end{cases}\Rightarrow}a=b=c}\)
Thay a=b=c vào P :
\(P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{a+a}{a}+\frac{a+a}{a}+\frac{a+a}{a}=6\)
Ta có :
\(\frac{a}{2}=\frac{b}{3};\frac{a}{4}=\frac{c}{9}\)
\(\Rightarrow\frac{a}{4}=\frac{b}{6}=\frac{c}{9}\)
\(\Rightarrow\frac{a^3}{64}=\frac{b^3}{216}=\frac{c^3}{729}\)
Áp dụng c/t tỉ lệ thức = nhau ta có :
\(\frac{a^3}{64}=\frac{b^3}{216}=\frac{c^3}{729}=\frac{a^3+b^3+c^3}{64+216+729}=\frac{-1009}{1009}=-1\)
- \(\frac{a^3}{64}=-1\Rightarrow a^3=-64\Rightarrow a=-4\)
- \(\frac{b^3}{216}=-1\Rightarrow b^3=-216\Rightarrow a=-6\)
- \(\frac{c^3}{729}=-1\Rightarrow c^3=-729\Rightarrow a=-9\)
Vậy a = -4 b = -6 c = -9
Ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\left(Đk:a;b;c\ne0\right)\)
Áp dụng tc của dãy tỉ số bằng nhau ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+2\right)}=\frac{1}{2}\)
=> \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{2}{1}=2\)
=> \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=2+2+2=6\)
a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b
=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1
= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1
= 2010.(1/b+c + 1/c+a + 1/a+b) - 3
= 2010.1/3 - 3 = 667
Vậy S = 667
Tk mk nha
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\Leftrightarrow a^3=c^3=b^3\)
Ta có : \(a^3=b^3=c^3=abc\)
\(\frac{a^3}{abc}=\frac{abc}{abc}=1\Leftrightarrow\frac{a^3+b^3+c^3}{3abc}=\frac{3abc}{3abc}=1\)
Vậy \(P=1\)