\(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
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,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
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 \(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
\(\Rightarrow A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(z+x\right)\left(z+y\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)}{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
Chứng minh rằng: \(\frac{x^2-y^2}{\left(z+x\right)\left(z+y\right)}+\frac{y^2-z^2}{\left(x+y\right)\left(x+z\right)}+\frac{z^2-x^2}{\left(y+z\right)\left(y+x\right)}=\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x}\)
103,CM:\(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{zx}+\frac{z\left(y-x\right)}{xy}}=x+y+z\)
\(\Leftrightarrow\) \(\frac{\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)}\)\(+\frac{\left(y-x\right)-\left(y-z\right)}{\left(y-z\right)\left(y-x\right)}+\frac{\left(z-y\right)-\left(z-x\right)}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
\(\Leftrightarrow\)\(\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
\(\Leftrightarrow\)\(\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}+\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
tự lm nốt ik
103,CM:\(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{zx}+\frac{z\left(y-x\right)}{xy}}=x+y+z\)
Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)
=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)
\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
Như vậy:
\(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)
\(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)
\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
thực hiện phép tính
\(M=\frac{2}{x-y}-\frac{2}{y-z}-\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
câu 1: giải hệ phương trình
\(\left(x+y\right)^2+\left(y+z\right)^4+....+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
\(\left(xy\right)^2+2\left(yz\right)^4+....+100\left(zx\right)^{100}=-[\left(x+y+z\right)+2\left(yz+zx+xy\right)+......+99\left(x+y+z\right)]\)\(\left(\frac{1}{x}+\frac{1}{y}\right)^2+\left(\frac{1}{y^2}+\frac{1}{z^2}\right)^2+...+\left(\frac{1}{x^{99}}+\frac{1}{z^{99}}\right)^2=-\frac{1}{\left(xy\right)^2+2\left(yz\right)^2+.....+99\left(zx\right)^2}\)
tìm x,y,z
Đúng là chơi lừa bịp thực sự bài này rất dễ đây là cách giải:
ta có: \(\left(x+y\right)^2+\left(y+z\right)^4+.....+\left(x+z\right)^{100}\ge0\)còn \(-\left(y+z+x\right)\le0\) nên phương trình 1 vô lý
tương tự chứng minh phương trinh 2 và 3 vô lý
vậy \(\hept{\begin{cases}x=\varnothing\\y=\varnothing\\z=\varnothing\end{cases}}\)
thực sự bài này mới nhìn vào thì đánh lừa người làm vì các phương trình rất phức tạp nhưng nếu nhìn kĩ lại thì nó rất dễ vì các trường hợp đều vô nghiệm
\(\left(x+y\right)^2+\left(y+z\right)^4+...+\left(x+z\right)^{100}=-\left(y+z+x\right)\)
Đặt : \(A=\left(x+y\right)^2+\left(y+z\right)^4+...+\left(x+z\right)^{100}\)
Ta dễ dàng nhận thấy tất cả số mũ đều chẵn
\(=>A\ge0\)(1)
Đặt : \(B=-\left(y+z+x\right)\)
\(=>B\le0\)(2)
Từ 1 và 2 \(=>A\ge0\le B\)
Dấu "=" xảy ra khi và chỉ khi \(A=B=0\)
Do \(B=0< =>y+z+x=0\)(3)
\(A=0< =>\hept{\begin{cases}x+y=0\\y+z=0\\x+z=0\end{cases}}\)(4)
Từ 3 và 4 \(=>x=y=z=0\)
Vậy nghiệm của pt trên là : {x;y;z}={0;0;0}
Đặt :\(\left(xy\right)^2+2\left(yz\right)^4+...+100\left(zx\right)^{100}=A\)
Ta thấy các số mũ đều chẵn
Nên \(A\ge0\left(1\right)\)
Đặt : \(-\left[\left(x+y+z\right)+2\left(yz+zx+xy\right)+...+99\left(x+y+z\right)\right]=B\)
Vì có dấu âm ở trước VT
Nên \(B\le0\left(2\right)\)
Từ 1 và 2 <=> \(A=B=0\)
\(< =>x=y=z=0\)
Simplify:
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}=\frac{x^2+xy}{\left(x+y\right)\left(x+z\right)}-\frac{xy+yz}{\left(x+y\right)\left(x+z\right)}=\frac{x}{x+z}-\frac{y}{x+y}\)
Tương tự:\(\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}=\frac{y}{x+y}-\frac{z}{y+z};\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=\frac{z}{z+y}-\frac{x}{z+x}\)
Khi đó:
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=0\)