Cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). Tính giá trị của bt \(M=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
Cho x, y, z đôi một khác nhau và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\).Tính giá trị của biểu thức D=\(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+xz}+\dfrac{xy}{z^2+xy}\)
⇔xy+yz+zx=0
=yz/(x−y)(x−z)
Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)
Cho x, y, z đôi một khác nhau và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\).Tính giá trị của biểu thức: \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)
Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
⇒yz=−xy−zx⇒yz/x^2+2yz=yz/x^2+yz−xy−zx
=yz/(x−y)(x−z)
Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+xz=0\)
\(\Leftrightarrow yz=-xy-xz\)\(\Leftrightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\)
\(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Cho x,y,z đôi một khác nhau và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). Tính giá trị của biểu thức: \(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)=0. Tính giá trị của biểu thức A=\(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
Mong mọi người giúp đỡ
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
Cho x, y, z đôi một khác nhau và \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\) = 0
Tính giá trị của biểu thức: M = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
Giúp mk giải bài này với, khó quá :((
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
Cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
Tính \(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
\(x,y,z\ne0\)
-Ta c/m: -Với \(a+b+c=0\) thì: \(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0.\left(a^2+b^2+c^2-ab-bc-ca\right)=0\left(đpcm\right)\)
-Quay lại bài toán:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)
1.a.rút gọn biểu thức M = \(\dfrac{\text{1}}{\text{(x - y)(z² + yz - x² - xy)}}-\dfrac{\text{1}}{\text{(y - z)(x² + xz - y² -yz)}}+\dfrac{\text{1}}{\text{(z - x)(y² + xy - z² - xz)}}\)
b. tính giá trị của M tại x = y = z = 2015
\(\dfrac{1}{\left(x-y\right)\left(z^2+yz-x^2-xz\right)}=\dfrac{1}{\left(x-y\right)\left[\left(z-x\right)\left(z+x\right)+y\left(z-x\right)\right]}=\dfrac{1}{\left(z-x\right)\left(x-y\right)\left(x+y+z\right)}\)
Tương tự: \(\dfrac{1}{\left(y-z\right)\left(x^2+xz-y^2-yz\right)}=\dfrac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}\)
\(\dfrac{1}{\left(z-x\right)\left(y^2+xy-z^2-xz\right)}=\dfrac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}\)
\(\Rightarrow M=\dfrac{y-z-z+x-x+y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}\\ M=\dfrac{2}{\left(x-y\right)\left(z-x\right)\left(x+y+z\right)}\)
\(a,\) Bổ sung điều kiện: \(x\ne y\ne z\)
\(b,\) Đề bài ko thỏa mãn điều kiện nên không tính đc M
cho x,y, z khác 0 thỏa mãn \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=0.\)tính \(\dfrac{x^2}{yz}+\dfrac{z^2}{xy}+\dfrac{y^2}{xz}\)
Do \(xyz\ne0\) ta có:
\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)
Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)
Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
Cho x, y, z khác 0, \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). Chứng minh rằng: \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Trước hết, ta đi chứng minh một bổ đề sau: Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\). Thật vậy, ta phân tích
\(P=a^3+b^3+c^3-3abc\)
\(P=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(P=\left(a+b+c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(P=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).
Hiển nhiên nếu \(a+b+c=0\) thì \(P=0\) hay \(a^3+b^3+c^3=3abc\), bổ đề được chứng minh.
Do \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) nên áp dụng bổ đề, ta được \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\).
Vì vậy \(\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\) \(=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\) \(=xyz.\dfrac{3}{xyz}=3\). Ta có đpcm