Tìm x, y, z biết: \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{-2}{x^2}=\frac{-xz^2-yz^2}{z^2+1}\left(x\ne0\right)\)
Chmr nếu:
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}vớix\ne y,yz\ne1,xz\ne1,x\ne0,y\ne0,z\ne0\)
thì: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(chox,y,z\ne0;x\ne y;\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}CM:x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
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vô phần thống kê hỏi đáp của mình để coi hình nhé
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)
\(\Leftrightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+xy^3z+x^2z-x^2yz^2=0\)
\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)-xyz^2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right]=0\)
\(\Leftrightarrow xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2=0\left(x\ne y\Rightarrow x-y\ne0\right)\)
\(\Leftrightarrow xy+yz+xz=xyz\left(x+y\right)+xyz^2\)
\(\Leftrightarrow\frac{ay+yz+xz}{xyz}=\frac{xyz\left(x+y\right)+xyz^2}{xyz}\left(xyz\ne0\right)\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z\)
Chmr nếu:
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) với \(x\ne y,yz\ne1,xz\ne1,x\ne0,y\ne0,z\ne0\)
thì: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng t/c dãy tỉ số bằng nhau có:
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x-xyz-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
=> \(\frac{x^2-yz}{x\left(1-yz\right)}=x+y+z\)
<=> \(\frac{x^2-yz}{x\left(1-yz\right)}-\frac{\left(x+y+z\right)x\left(1-yz\right)}{x\left(1-yz\right)}=0\)
<=> \(\frac{x^2-yz-\left(x^2+yx+zx\right)\left(1-yz\right)}{x\left(1-yz\right)}\)=0
<=> \(x^2-yz-x^2+x^2yz-xy+xy^2z-xz+xyz^2=0\)
<=> \(-yz-xy-xz+xyz\left(x+y+z\right)\)=0
<=> \(xyz\left(x+y+z\right)=yz+xy+xz\)
<=>\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)( chia cả hai vế cho xyz với x,y,z khác 0)
Cho \(\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\left(x,y,z\ne0\right)\)
Tìm: \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\left(x,y,z\ne0\right).\)
Tính \(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
Câu hỏi của Vũ Thảo Vy - Toán lớp 8 - Học toán với OnlineMath tham khảo
Cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}=0\left(x,y,z\ne0\right).\)
Tinh
\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
vt sai đề nâk
từ gt=> xy+yz+xz=0
áp dụng bdt bunhia
=> A>=0
dấu= xr khi x=y=z
-> dấu = k xr
..........
hoặc:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{\Rightarrow1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)
chứng minh nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)với x\(\ne y,xyz\ne0,yz\ne1,xz\ne1\) thì xy+yz+zx=xyz(x+y+z)
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
103,CM:\(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{zx}+\frac{z\left(y-x\right)}{xy}}=x+y+z\)
chứng minh rằng nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\) với \(x\ne y,xyz\ne0,yz\ne1,xz\ne1\)thì xy+yz+xz=xyz(x+y+z)
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)
\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)
\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)
\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)
\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)
\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)
Mà \(x\ne y\) nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)
\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)
Đpcm
Từ gt ta có : (x2 - yz)y(1 - yz) = (y2 - xz)x(1 - yz)
=> 0 = VT - VP = (x2y - x3yz - y2z - xy2z2) - (xy2 - xy3z - x2z - x2yz2) = xy(x - y) - xyz(x2 - y2) + z(x2 - y2) + xyz2(y - x)
= (x - y)[xy - xyz(x + y) + z(x + y) - xyz2] = (x - y)(xy + yz + xz - xyz(x + y + z)]
Vì\(x\ne y\Rightarrow x-y\ne0\) nên xy + yz + xz - xyz(x + y + z) = 0 => xy + yz + xz = xyz(x + y + z)
Bạn ko hiểu chỗ nào thì hỏi mình nhé!