Phát biểu nào là sai ?

A. Nếu AB\overrightarrow{AB} = AC\overrightarrow{AC} thì AB\left|\overrightarrow{AB}\right| = AC\left|\overrightarrow{AC}\right|

B. AB\overrightarrow{AB} = CD\overrightarrow{CD} thì A,B,C,D thẳng hàng

C. Nếu 3. AB\overrightarrow{AB} + 7 . AC\overrightarrow{AC} = 0\overrightarrow{0} thì A,B,C thẳng hàng

D. AB\overrightarrow{AB} - CD\overrightarrow{CD} = DC\overrightarrow{DC} - BA\overrightarrow{BA}

Các câu hỏi liên quan

Đề: Cho {x,y,z>0x+yz\left\{{}\begin{matrix}x,y,z>0\\x+y\le z\end{matrix}\right. tìm Min của (x2+y2+z2)(1x2+1y2+1z2)\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right) Làm thế này không biết đúng ko

Ta có :A= (x2+y2+z2)(1x2+1y2+1z2)=3+x2y2+y2x2+z2x2+x2z2+z2y2+y2z2\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{z^2}{x^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{y^2}+\dfrac{y^2}{z^2}

=> A =3+(x2y2+y2x2)+(x2z2+z216x2)+(y2z2+z216y2)+1516(z2x2+z2y2)=3+\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)

Áp dụng BĐT Cauchy ta có

A3+2+12+12+1516(z2x2+z2y2)=6+1516(z2x2+z2y2)A\ge3+2+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)=6+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)

Do x+yzxz+yz1x+y\le z\Rightarrow\dfrac{x}{z}+\dfrac{y}{z}\le1 ; Đặt u=xzu=\dfrac{x}{z}; v=yzv=\dfrac{y}{z}

z2x2+z2y2=1u2+1v22uv2(u+v)24214=8\Rightarrow\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}=\dfrac{1}{u^2}+\dfrac{1}{v^2}\ge\dfrac{2}{uv}\ge\dfrac{2}{\dfrac{\left(u+v\right)^2}{4}}\ge\dfrac{2}{\dfrac{1}{4}}=8

A6+1516.8=272\Rightarrow A\ge6+\dfrac{15}{16}.8=\dfrac{27}{2} Vậy minA = 272\dfrac{27}{2} khi x=y=z2x=y=\dfrac{z}{2}