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\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
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\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
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\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
a/ +) \(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\)\(\left(1\right)\)
+) \(\dfrac{y}{3}=\dfrac{z}{5}\Leftrightarrow\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\\\dfrac{y}{12}=3\\\dfrac{z}{20}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
Vậy ..
b/ \(2x=3y=5z\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)
\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Theo t/c dãy tỉ số bằng nhau tcos :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=5\\\dfrac{y}{10}=5\\\dfrac{z}{6}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=75\\y=50\\z=30\end{matrix}\right.\)
Vậy..
c/ tương tự
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
Cách khác:
Áp dụng BĐT AM-GM:
\(\frac{a}{b^2}+\frac{1}{a}\geq 2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)
\(\frac{b}{c^2}+\frac{1}{b}\geq 2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)
\(\frac{c}{a^2}+\frac{1}{c}\geq 2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)
Cộng theo vế và rút gọn:
\(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
a, \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{x^2+y^2}{4+16}=\dfrac{2000}{20}=100\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=100.4=400\\y^2=100.16=1600\\z^2=100.25=2500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm20\\y=\pm40\\z=\pm50\end{matrix}\right.\)
Do \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}x=20\\y=40\\z=50\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-20\\y=-40\\z=-50\end{matrix}\right.\)
Vậy ...
b, \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.3=3\\z-3=1.4=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Vậy ...
c, \(x-z=-2\Rightarrow x+2=z\)
Do đó \(y.z=12\Leftrightarrow y.\left(x+2\right)=12\Rightarrow xy+2y=12\Rightarrow6+2y=12\)
\(\Rightarrow y=3\Rightarrow x.3=6\Rightarrow x=2\Rightarrow2-z=-2\Rightarrow z=4\)
Vậy x=2; y=3; z=4
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
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