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a) \(A\left(x\right)=x^2-10x+25\)
\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)
b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)
\(\Rightarrow B\left(x\right)=5x^2+5x\)
\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)
c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)
\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)
d) Nghiệm của B(x)
\(\Leftrightarrow B=0\)
\(\Leftrightarrow5x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)
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a, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)
Áp dụng tính chất của day tỉ số bằng nhau ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}=\dfrac{3a+b}{3c+d}\)
\(=>\dfrac{a}{c}=\dfrac{3a+b}{3c+d}=>\dfrac{a}{3a+b}=\dfrac{c}{3c+d}=>\left(đpcm\right)\)
Bài 1:
Ta có:\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}=\dfrac{3a+b}{3c+d}\)
⇒\(\dfrac{a}{c}=\dfrac{3a+b}{3c+d}\Rightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
Vậy từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)(ĐPCM)
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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 ...