Cộng các phân thức đại số :
\(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
Cộng các phân thức đại số sau vào với nhau:
\(\frac{1}{\left(y-z\right)\left(x^2+xz-y^2-yz\right)}+\frac{1}{\left(z-x\right)\left(y^2+xy-z^2-zx\right)}+\frac{1}{\left(x-y\right)\left(z^2+yz-x^2-xy\right)}\)
Giúp mình với các bạn
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,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
1.Tính:
\(x:\frac{x-1}{2}-\frac{\left(x-1\right)\left(x^2+4x+1\right)}{2x^2+2x}.\frac{-4x}{\left(x-1\right)^2}-\frac{4x^2}{x^2-1}\)
2.Chứng minh đẳng thức sau( giả sử đẳng thức có nghĩa):
\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
Các bạn giúp mình với!
Cho các số x,y,z là các số phức phân biệt sao cho \(y=tx+\left(1-t\right)z,t\in\left(0,1\right)\)
Chứng minh rằng :
\(\frac{\left|z\right|-\left|y\right|}{\left|z-y\right|}\ge\frac{\left|z\right|-\left|x\right|}{\left|z-x\right|}\ge\frac{\left|y\right|-\left|x\right|}{\left|y-x\right|}\)
Từ hệ thức :
\(y=tx+\left(1-t\right)z\)
Bất đẳng thức
\(\frac{\left|z\right|-\left|y\right|}{\left|z-y\right|}\ge\frac{\left|z\right|-\left|x\right|}{\left|z-x\right|}\)
Trở thành :
\(\left|z\right|-\left|y\right|\ge t\left(\left|z\right|-\left|x\right|\right)\)
hay
\(\left|y\right|\le\left(1-t\right)\left|z\right|+t\left|x\right|\)
Vận dụng bất đẳng thức tam giác cho
\(y=\left(1-t\right)x+tx\) ta có kết quả
Bất đẳng thức thứ hai, được chứng minh tương tự bởi
\(y=tx+\left(1-t\right)z\)
tương đương với :
\(y-x=\left(1-t\right)\left(z-x\right)\)
Rút gọn phân thức
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}\)+\(\frac{1}{y\left(y-z\right)\left(y-x\right)}\)+\(\frac{1}{z\left(z-y\right)\left(z-x\right)}\)
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 \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
\(\Rightarrow A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(z+x\right)\left(z+y\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)}{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
cho \(x;y;z>0\)
\(xy+yz+xz=xyz\)
và \(\left(x+y\right)\left(\frac{1}{z}+\frac{1}{xy}\right)+\left(y+z\right)\left(\frac{1}{x}+\frac{1}{yz}\right)+\left(x+z\right)\left(\frac{1}{y}+\frac{1}{xz}\right)=1\)
tính giá trị của biểu thức
\(A=\sqrt{\frac{\left(2x+yz\right)\left(2y+xz\right)}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{\left(2y+xz\right)\left(2z+xy\right)}{\left(x+z\right)\left(x+y\right)}}+\sqrt{\frac{\left(2z+xy\right)\left(2x+yz\right)}{\left(x+y\right)\left(y+z\right)}}\)
Xem lại cái đề đi Tuyển. Hình như giá trị nhỏ nhất của cái biểu thức dưới còn lớn hơn là 1 thì làm sao bài đó có giá trị x, y, z thỏa được mà bảo tính A.
Cộng trừ phân số
1)\(x+2+\frac{3}{x-2}\)
2)\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
1)
\(x+2+\frac{3}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4}{x-2}+\frac{3}{x-2}\)
\(=\frac{x^2-4+3}{x-2}\)
\(=\frac{x^2-1}{x-2}\)
2)
\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{y^2\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{\left(x^2-xy-xz+yz\right)\left(y-z\right)}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-xy^2-xyz+y^2z-x^2z+xyz+xz^2-yz^2}\)
\(=\frac{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}{x^2y-x^2z-xy^2+y^2z+xz^2-yz^2}\)
\(=1\)
\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\left(\frac{x^2}{x-z}-\frac{y^2}{y-z}\right)+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\left(\frac{x^2\left(y-z\right)}{\left(x-z\right)\left(y-z\right)}-\frac{y^2\left(x-z\right)}{\left(y-z\right)\left(x-z\right)}\right)+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\frac{x^2y-x^2z-xy^2+y^2z}{\left(x-z\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\frac{xy\left(x-y\right)-z\left(x^2-y^2\right)}{\left(x-z\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\frac{xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)}{\left(x-z\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\frac{\left(x-y\right)\left(xy-z\left[x+y\right]\right)}{\left(x-z\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}\times\frac{\left(x-y\right)\left(xy-xz-zy\right)}{\left(x-z\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{xy-xz-zy+z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{y\left(x-z\right)-z\left(x-z\right)}{y\left(x-z\right)-z\left(x-z\right)}\)
= 1
Thu gọn đa thức
\(\frac{1}{x\left(x-y\right)\left(y-z\right)}+\frac{1}{y\left(y-z\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)