\(\left|x+y\right|< \left|x\right|+\left|y\right|\)
Hãy so sánh 2 biểu thức sau , biết rằng \(x;y\in Z\):
\(\left[\left(x+y\right)\left(x+y\right)+\left(x-y\right)\left(x-y\right)\right]\left[\left(x+y\right)\left(x+y\right)-\left(x-y\right)\left(x-y\right)\right]\)
và
\(\left[\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x-y\right)\right]\left[\left(x+y\right)\left(x-y\right)-\left(x+y\right)\left(x-y\right)\right]\)
j toàn chữ là chữ vậy trời , làm tớ hoa hết cả mắt !!@@@@@@@
+Biểu thức thứ nhất
\(=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left[\left(x+y\right)^2-\left(x-y\right)^2\right]=\left(x+y\right)^4-\left(x-y\right)^4\)
+Biểu thức thứ hai
\(=\left(x^2-y^2+x^2-y^2\right).0=0\)
2 biểu thức này khác nhau.
ta có : \(x^2+1=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự ta đc \(y^2+1=\left(y+x\right)\left(y+z\right)\)
\(z^2+1=\left(z+x\right)\left(z+y\right)\)
ĐẶt \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
\(\Rightarrow A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(z+x\right)\left(z+y\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)}{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x+xy-y+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y-x^2-x-xy+y-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\\left(y+1\right)^2=\left(y-1\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1=y^2-3y+2+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1-y^2+3y-2-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x=2\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\-1+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\3y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy hpt trên có nghiệm duy nhất (x;y) = (-1; \(\dfrac{1}{3}\))
Chúc bn học tốt!
\(\left\{{}\begin{matrix}x^2+xy-x-y=x^2+x-xy-y+2xy+2\\y^2-xy-x+y=y^2+xy-2y-2x-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x=2\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2+xy-x-y=x^2+xy+x-y+2\\y^2+y-xy-x=y^2-xy-2y-2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x=2\\y-x+2y+2x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y=-x=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
a) \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)
b) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
c) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
a) CMR : \(\frac{\left|x\right|}{\left|y\right|+2}+\frac{\left|y\right|}{\left|x\right|+2}\ge\frac{\left|x\right|+\left|y\right|}{\left|x\right|+\left|y\right|+2}\)
b) CMR \(\frac{\left|x\right|}{\left|y\right|+2}+\frac{\left|y\right|}{\left|x\right|+2}\ge\frac{\left|x+y\right|}{\left|x+y\right|+2}\)
Cộng các phân thức :
a) \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)
b) \(\dfrac{4}{\left(y-x\right)\left(z-x\right)}+\dfrac{3}{\left(y-x\right)\left(y-z\right)}+\dfrac{3}{\left(y-z\right)\left(x-z\right)}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
Chứng minh đẳng thức:
a) \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}+\dfrac{z}{\left(y-z\right)\left(z-x\right)}+\dfrac{x}{\left(z-x\right)\left(x-y\right)=0}\)
b) \(\dfrac{x^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)=1}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Tính:
a) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
b) \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-x\right)\left(y-z\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
d) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
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d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)