\(\overrightarrow{c}=2\left(2;1\right)+3\left(3;-2\right)=\left(4+9;2-6\right)=\left(13;-4\right)\)

a: \(\overrightarrow{MN}+\overrightarrow{NP}+\overrightarrow{PQ}\)

\(=\overrightarrow{MP}+\overrightarrow{PQ}\)

\(=\overrightarrow{MQ}\)

Ta có: \(\left\{{}\begin{matrix}\overrightarrow{a}=m\overrightarrow{u}+\overrightarrow{v}=\left(4m+1;m+4\right)\\\overrightarrow{b}=\overrightarrow{i}+\overrightarrow{j}=\left(1;1\right)\end{matrix}\right.\)

Yêu cầu bài toán <=> cos\(\left(\overrightarrow{a};\overrightarrow{b}\right)\)=cos45^o =\(\dfrac{\sqrt{2}}{2}\)

<=> \(\dfrac{\left(4m+1\right)+\left(m+4\right)}{\sqrt{2}\sqrt{\left(4m+1\right)^2+\left(m+4\right)^2}}=\dfrac{\sqrt{2}}{2}\)

<=> \(\dfrac{5\left(m+1\right)}{\sqrt{2}\sqrt{17m^2+16+17}}=\dfrac{\sqrt{2}}{2}\)

<=> \(5\left(m+1\right)=\sqrt{17m^2+16m+17}\)  <=>\(\left\{{}\begin{matrix}m+1\ge0\\25m^2+50m+25=17m^2+16m+17\end{matrix}\right.\)

<=> m=\(-\dfrac{1}{4}\)

\(BA=\sqrt{12^2+8^2}=4\sqrt{13}=4BC\)

Do \(C\in\) tia BA \(\Rightarrow\overrightarrow{BC}=\frac{1}{4}\overrightarrow{BA}\)

Gọi \(C\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BC}=\left(x+11;y+25\right)\\\overrightarrow{BA}=\left(12;8\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+11=\frac{1}{4}.12\\y+25=\frac{1}{4}.8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-23\end{matrix}\right.\) \(\Rightarrow C\left(-8;-23\right)\)

2:

a: =>1/3x=16+1/4-13-1/4=3

=>x=9

b: =>4,5-2x=11/14:(-11/7)=-1/2=-0,5

=>2x=5

=>x=2,5

3:

Số học sinh giỏi là 12*5/6=10 bạn

Số học sinh trung bình là 10*1,4=14 bạn

Số học sinh của lớp là 10+12+14=36 bạn

\(d'=T_{\overrightarrow{v}}\left(d\right)\)

Ta có: \(\left\{{}\begin{matrix}x'=x+a\\y'=y+b\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=x'-a=x'-3\\y=y'-b=y'-4\end{matrix}\right.\)

Thay vào pt \(\left(d\right):x+y-6=0\) ta đc:

\(\Rightarrow\left(x'-3\right)+\left(y'-4\right)-6=0\)

\(\Rightarrow x'+y'-13=0\)

Vậy \(\left(d'\right):x+y-13=0\)

Bằng vec tơ gì bạn ơi ?