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Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x+y}{2012}=\dfrac{xy}{2013}=\dfrac{x-y}{2014}=\dfrac{x+y+x-y}{2012+2014}=\dfrac{2x}{4026}=\dfrac{x}{2013}\)
\(hay:\dfrac{xy}{2013}=\dfrac{x}{2013}\Rightarrow xy=x\Rightarrow y=1\)
Ta có:
\(\dfrac{x+y}{2012}=\dfrac{x}{2013}\) (c/m trên)
\(\Rightarrow\left(x+y\right)2013=2012x\\ hay:\left(x+1\right)2013=2012x\\ \Rightarrow2013x+2013=2012x\\ \Rightarrow2013x-2012x=-2013\\ \Rightarrow x=-2013\)
Vậy: x=-2013
phần a
vì x/2= y/3
y/5= z/4
=>x/2 nhân 1.5 = y/3 nhân 1/5
=> y/5 nhân 1/3 = z/4 nhân 1/3
=>x/10 = y/15 (1)
=>y/15 = z/12 (2)
Từ (1) , (2) ta có :
x/10 = y/15 = z/12
áp dụng t/c......
=>x/10 = y/15 = z/12
=>x+y+z/10+15+12
=> -49/37
b lm tiếp bc tiếp theo nhé✔
Vì mk cmt đầu tiên lên b tích dùm m☢
Lời giải:
\(\frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}\)
\(\Rightarrow (\frac{x+y}{y+z})^4=(\frac{y+z}{z+t})^4=(\frac{z+t}{t+x})^4=(\frac{t+x}{x+y})^4=\frac{x+y}{y+z}.\frac{y+z}{z+t}.\frac{z+t}{t+x}.\frac{t+x}{x+y}=1\)
\(\Rightarrow \left[\begin{matrix} \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=1\\ \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=-1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=y=z=t\\ x+y+z+t=0\end{matrix}\right.\)
Nếu $x=y=z=t$ thì:
\(A=\left(\frac{y+z}{x+t}\right)^{2013}+\left(\frac{y+t}{x+y}\right)^{2014}=\left(\frac{x+x}{x+x}\right)^{2013}+\left(\frac{x+x}{x+x}\right)^{2014}=1+1=2\in\mathbb{Z}\)
Nếu $x+y+z+t=0$ thì:
\(y+z=-(x+t); y+t=-(x+y)\)
\(\Rightarrow A=(-1)^{2013}+(-1)^{2014}=(-1)+1=0\in\mathbb{Z}\)
Vậy biểu thức $A$ luôn có giá trị nguyên.
Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Ta có :
x-y-z=0 => y+z=x (*(
Thay (*) và đa thức M ta có :
M=\(xyz-xy^2-xz^2=\left(y+z\right)yz-\left(y+z\right)y^2-\left(y+z\right)z^2\)
=\(y^2z+yz^2-y^3-zy^2-z^2y-z^3\)
=\(\left(y^2z-y^2z\right)-\left(z^2y-z^2y\right)-\left(y^3+z^3\right)\)
=\(-\left(y^3+z^3\right)\)
Mà \(-\left(y^3+z^3\right)\) là số đối của \(\left(y^3+z^3\right)\) nên M và N là 2 đa thức đối nhau.
Câu 1 :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.......+\dfrac{1}{2012}\right)\)=\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\)=P
Vậy S=P
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)
\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)
\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)
\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)
\(\Rightarrow x-2015=0\)
\(x=0+2015\)
\(x=2015\)
a, ta co:
x-2/4=-16/2-x
=>(x-2)(2-x)=(-16).4
lai co: x-2/2-x=-1
=>x-2=(-1).(2-x)
a, ta co:
x-2/4=-16/2-x
=>(x-2)(2-x)=(-16).4 (1)
lai co: x-2/2-x=-1
=>x-2=(-1).(2-x) (2)
thay(2) vao(1) ,ta co:
(2-x)^2=-64
.........(tu lam tiep nha)
1/ a, Ta có :
\(x-2y+3z=35\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}=\dfrac{7}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{7}{2}\Leftrightarrow x=\dfrac{21}{2}\\\dfrac{x}{4}=\dfrac{7}{2}\Leftrightarrow y=14\\\dfrac{z}{5}=\dfrac{7}{2}\Leftrightarrow z=\dfrac{35}{2}\end{matrix}\right.\)
Vậy ..