Question

Bài 1:Cho x,y,z là 3 số khác 0.thỏa mãn \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)\(1\)

TÍNH GT BT

\(A=\left(x^{25}+y^{25}\right)\left(y^3+z^3\right)\left(x^{2019}+z^{2019}\right)\)

Answer 1

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)

\(\Rightarrow A=0\)

Answer 2

Sai đề không

Answer 3

Sai đề thật, cái biểu thức trên 100% lớn hơn hoặc = 9, lấy đâu ra =1

Answer 4

mk viết lộn 

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\)\(1\)

Answer 5

Phạm Minh Đức bằng 1 thì làm được chứ bằng 11 thì có lẽ sai đề

Answer 6

uk mk viết nhầm

Answer 7

Ủa, sao mk chứng minh đc cái biểu thức kia lớn hơn hoặc bằng 9 nhỉ

Answer 8

1  chứ không phải 11 nha :))

\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1=\frac{x+y+z}{x+y+z}\)

\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\left(\text{Vì }x+y+z\ne0\right)\)

\(\Rightarrow\frac{\left(x+y+z\right).\left(xy+yz+xz\right)}{xyz\left(x+y+z\right)}-\frac{xyz}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow x^2y+xyz+x^2z+xy^2+zy^2+xyz+xyz+yz^2+xz^2-xyz=0\)

\(\Rightarrow\left(x^2y+xyz\right)+\left(x^2z+xz^2\right)+\left(xyz+yz^2\right)+\left(xy^2+zy^2\right)=0\)

\(\Rightarrow xy.\left(x+z\right)+xz.\left(x+z\right)+yz.\left(x+z\right)+y^2.\left(x+z\right)=0\)

\(\Rightarrow\left(x+z\right).\left(xy+xz+yz+y^2\right)=0\)

\(\Rightarrow\left(x+z\right).\left[y.\left(x+y\right)+z.\left(x+y\right)\right]=0\)

\(\Rightarrow\left(x+z\right).\left(x+y\right).\left(y+z\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-z\\x=-y\end{cases}\text{hoặc }y=-z}\)

\(A=\left(x^{25}-x^{25}\right).\left(y^3+z^3\right).\left(x^{2019}+z^{2019}\right)=0\)

Answer 9

cho t copy tý nha Boul:

"1  chứ khong phải 11 nha :))

\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1=\frac{x+y+z}{x+y+z}\)

\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\left(\text{Vì }x+y+z\ne0\right)\)

\(\Rightarrow\frac{\left(x+y+z\right).\left(xy+yz+xz\right)}{xyz\left(x+y+z\right)}-\frac{xyz}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow x^2y+xyz+x^2z+xy^2+zy^2+xyz+xyz+yz^2+xz^2-xyz=0\)

\(\Rightarrow\left(x^2y+xyz\right)+\left(x^2z+xz^2\right)+\left(xyz+yz^2\right)+\left(xy^2+zy^2\right)=0\)

\(\Rightarrow xy.\left(x+z\right)+xz.\left(x+z\right)+yz.\left(x+z\right)+y^2.\left(x+z\right)=0\)

\(\Rightarrow\left(x+z\right).\left(xy+xz+yz+y^2\right)=0\)

\(\Rightarrow\left(x+z\right).\left[y.\left(x+y\right)+z.\left(x+y\right)\right]=0\)

\(\Rightarrow\left(x+z\right).\left(x+y\right).\left(y+z\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-z\\x=-y\end{cases}\text{hoặc }y=-z}\)

\(A=\left(x^{25}-x^{25}\right).\left(y^3+z^3\right).\left(x^{2019}+z^{2019}\right)=0\)"

Nguồn: Boul đz =)