Câu 1 :

a) \(\overrightarrow{AD}+\overrightarrow{CB}=\overrightarrow{AB}-\overrightarrow{DC}\)

\(\Leftrightarrow\overrightarrow{AD}-\overrightarrow{AB}=-\left(\overrightarrow{DC}+\overrightarrow{CB}\right)\)

\(\Leftrightarrow\overrightarrow{BD}=-\overrightarrow{DB}\left(đúng\right)\)

\(\Rightarrowđpcm\)

b) Ta có \(G\) là trọng tâm \(\Delta ABC\left(gt\right)\)

\(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{AA}+\overrightarrow{AB}+\overrightarrow{AC}=3\overrightarrow{AG}\)

\(\Leftrightarrow\overrightarrow{0}+\overrightarrow{AB}+\overrightarrow{AC}=3\overrightarrow{AG}\)

\(\Leftrightarrow\overrightarrow{AG}=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(đpcm\right)\)

Câu 2 :

Chiều cao \(H\) của tháp là :

\(tan\left(23,6^o+15,9^o\right)=\dfrac{H}{200}\)

\(\Leftrightarrow tan39,5^o=\dfrac{H}{200}\)

\(\Leftrightarrow H=200.tan39,5^o\approx165\left(m\right)\)

a:

A(2;4); B(-3;1); C(3;-1); D(x;y)

\(\overrightarrow{AB}=\left(-3-2;1-4\right)\)

=>\(\overrightarrow{AB}=\left(-5;-3\right)\)

\(\overrightarrow{DC}=\left(3-x;-1-y\right)\)

ABCD là hình bình hành

=>\(\overrightarrow{AB}=\overrightarrow{DC}\)

=>\(\left\{{}\begin{matrix}3-x=-5\\-1-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=2\end{matrix}\right.\)

vậy: D(8;2)

b: Tọa độ trọng tâm G của ΔABC là:

\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{2+\left(-3\right)+3}{3}=\dfrac{2}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{4+1+\left(-1\right)}{3}=\dfrac{4}{3}\end{matrix}\right.\)

vậy: \(G\left(\dfrac{2}{3};\dfrac{4}{3}\right)\)

c:

Gọi H(x;y) là tọa độ trực tâm của ΔABC

A(2;4); B(-3;1); C(3;-1); H(x;y)

\(\overrightarrow{AH}=\left(x-2;y-4\right);\overrightarrow{BC}=\left(6;-2\right)\)

\(\overrightarrow{BH}=\left(x+3;y-1\right);\overrightarrow{AC}=\left(1;-5\right)\)

H là trực tâm của ΔABC nên AH\(\perp\)BC; BH\(\perp\)AC

=>\(\left\{{}\begin{matrix}\overrightarrow{AH}\cdot\overrightarrow{BC}=0\\\overrightarrow{BH}\cdot\overrightarrow{AC}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\left(x-2\right)+\left(-2\right)\left(y-4\right)=0\\1\left(x+3\right)+\left(-5\right)\left(y-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x-12-2y+8=0\\x+3-5y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=12-8=4\\x-5y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-y=2\\x-5y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{7}\\y=\dfrac{13}{7}\end{matrix}\right.\)

vậy: H(9/7;13/7)

Gọi O(x;y) là tọa độ tâm đường tròn ngoại tiếp ΔABC

O(x;y); A(2;4); B(-3;1); C(3;-1)

\(OA^2=\left(2-x\right)^2+\left(4-y\right)^2=\left(x-2\right)^2+\left(y-4\right)^2\)

\(OB^2=\left(-3-x\right)^2+\left(1-y\right)^2=\left(x+3\right)^2+\left(y-1\right)^2\)

\(OC^2=\left(3-x\right)^2+\left(-1-y\right)^2=\left(x-3\right)^2+\left(y+1\right)^2\)

Vì O là tâm đường tròn ngoại tiếp ΔABC nên OA=OB=OC

=>\(\left\{{}\begin{matrix}\left(x-2\right)^2+\left(y-4\right)^2=\left(x+3\right)^2+\left(y-1\right)^2\\\left(x+3\right)^2+\left(y-1\right)^2=\left(x-3\right)^2+\left(y+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-4x+4+y^2-8y+16=x^2+6x+9+y^2-2y+1\\x^2+6x+9+y^2-2y+1=x^2-6x+9+y^2+2y+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-4x-8y+20=6x-2y+10\\6x-2y+10=-6x+2y+10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-10x+6y=-10\\12x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=-\dfrac{15}{4}\end{matrix}\right.\)

 vậy: O(-5/4;-15/4)

a) \(O\) là trọng tâm hình thoi \(ABCD\) (tính chất hình thoi)

\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}=4\overrightarrow{MO}\)

\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}=4\overrightarrow{MO}-\left(\overrightarrow{MC}+\overrightarrow{MD}\right)\)

\(\Rightarrow\) Sai

b) \(\overrightarrow{OA}-\overrightarrow{OB}=\overrightarrow{OC}-\overrightarrow{OD}\)

\(\Leftrightarrow\overrightarrow{BA}=\overrightarrow{DC}\left(vô.lý\right)\)

\(\Rightarrow\) Sai

c) \(\left(\widehat{\overrightarrow{AB};\overrightarrow{AD}}\right)=\widehat{BAD}=\dfrac{360^o-2.120^o}{2}=60^o\)

\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|^2=AB^2+AD^2+2.AB.AD.cos\left(\widehat{\overrightarrow{AB};\overrightarrow{AD}}\right)\)

\(\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AD}\right|^2=2AB^2+2AB^2.cos60^o\left(AB=AD\right)\)

\(\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AD}\right|^2=2AB^2+2AB^2.\dfrac{1}{2}=3AB^2\)

\(\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=AB\sqrt{3}=7\sqrt{3}\)

\(\Rightarrow\) Đúng

d) Đúng (\(O\) là trung điểm \(AC\) - tính chất hình thoi)

d, $1-2x=0\Leftrightarrow x=\dfrac{1}{2}$

$x^2-x-1=0\Leftrightarrow x=\dfrac{1\pm\sqrt 5}{2}$

Bảng xét dấu:

\begin{array}{|c|cc|}\hline x&-\infty&&\dfrac{1-\sqrt 5}{2}&&\dfrac{1}{2}&&\dfrac{1+\sqrt 5}{2}&&+\infty\\\hline 1-2x&&+&|&+&0&-&|&-&\\\hline x^2-x-1&&+&0&-&|&-&0&+&\\\hline f(x)&&+&0&-&0&+&0&-&\\\hline\end{array}

$f(x)>0\Rightarrow x\in\left(-\infty;\dfrac{1-\sqrt 5}{2}\right)\cup\left(\dfrac{1}{2};\dfrac{1+\sqrt 5}{2}\right)$

Vậy $ x\in\left(-\infty;\dfrac{1-\sqrt 5}{2}\right)\cup\left(\dfrac{1}{2};\dfrac{1+\sqrt 5}{2}\right)$

e, $x^2-1=0\Leftrightarrow x=\pm 1$

$x^2-3=0\Leftrightarrow x=\pm\sqrt 3$

$-3x^2+2x+8=0\Leftrightarrow \left[\begin{array}{1}x=2\\x=-\dfrac{4}{3}\end{array}\right.$

Bảng xét dấu:

$\begin{array}{|c|cc|}\hline x&-\infty&&-\sqrt 3&&-\dfrac{4}{3}&&-1&&1&&\sqrt 3&&2&&+\infty\\\hline x^2-1&&+&|&+&|&+&0&-&0&+&|&+&|&+&\\\hline x^2-3&&+&0&-&|&-&|&-&|&-&0&+&|&+&\\\hline -3x^2+2x+8&&-&|&-&0&+&|&+&|&+&|&+&0&-&\\\hline f(x)&&-&||&+&||&-&0&+&0&-&||&+&||&-&\\\hline\end{array}$

$f(x)>0\Rightarrow x\in\left(-\sqrt 3;-\dfrac{4}{3}\right)\cup (-1;1)\cup (\sqrt 3;2)$

Vậy $x\in\left(-\sqrt 3;-\dfrac{4}{3}\right)\cup (-1;1)\cup (\sqrt 3;2)$

Xét tam giác vuông \(IAB:\)

\(IB^2=IA^2+AB^2=\left(\dfrac{3a}{2}\right)^2+16a^2=\dfrac{73a^2}{4}\)

\(\Rightarrow IB=\dfrac{a\sqrt{73}}{2}\)

\(tan\widehat{AIB}=\dfrac{AB}{IA}=\dfrac{4a}{\dfrac{3a}{2}}=\dfrac{8}{3}\)

\(\Rightarrow tan\widehat{DIB}=tan\left(180^o-\widehat{AIB}\right)=-tan\widehat{AIB}=-\dfrac{8}{3}\)

\(1+tan^2\widehat{DIB}=\dfrac{1}{cos^2\widehat{DIB}}\)

\(\Rightarrow cos^2\widehat{DIB}=\dfrac{1}{1+tan^2\widehat{DIB}}=\dfrac{1}{1+\dfrac{64}{9}}=\dfrac{9}{73}\)

\(\Rightarrow cos\widehat{DIB}=-\dfrac{3}{\sqrt{73}}\) (\(\widehat{DIB}\) thuộc góc phần tư số \(II\))

\(\Rightarrow cos\left(\widehat{\overrightarrow{IB};\overrightarrow{ID}}\right)=cos\widehat{DIB}=-\dfrac{3}{\sqrt{73}}\)

\(\left(\overrightarrow{IA}+\overrightarrow{IB}\right).\overrightarrow{ID}=\overrightarrow{IA}.\overrightarrow{ID}+\overrightarrow{IB}.\overrightarrow{ID}\)

\(=IA.ID.cos180^o+IB.ID.cos\left(\widehat{\overrightarrow{IB};\overrightarrow{ID}}\right)\)

\(=\dfrac{3a}{2}.\dfrac{3a}{2}.\left(-1\right)+\dfrac{a\sqrt{73}}{2}.\dfrac{3a}{2}.\left(-\dfrac{3}{\sqrt{73}}\right)\)

\(=-\dfrac{9a^2}{4}-\dfrac{9a^2}{4}=-\dfrac{9a^2}{2}\)

Xem lại đề bài

Chọn gốc tọa độ \(O\left(0;0\right)\) tại đỉnh Parabol, \(\overrightarrow{Ox}\) hướng từ trái sang phải; \(\overrightarrow{Oy}\) hướng từ dưới lên trên

Đồ thị Parabol có dạng \(y=ax^2\left(P\right)\)

Tọa độ chân cổng \(\left(2,1;-4,41\right)\in\left(P\right)\Leftrightarrow-4,41=a.2,1^2\Leftrightarrow a=-1\)

\(\Rightarrow y=-x^2\)

Chiều rộng xe tải là \(2,4\left(m\right)\)

\(\Rightarrow\) Hoành độ 2 bánh xe là \(x=\pm\dfrac{2,4}{2}=\pm1,2\)

\(\Rightarrow\) Tung độ của xe 2 bánh xe chính là chiều cao tối đa của xe

\(\left|y\right|=\left|-1,2^2\right|=1,44\left(m\right)\)

Vậy anh Quang dự định mua xe tải có chiều cao tối đa để qua cổng là \(1,44\left(m\right)\)