Lời giải:
Trên tia đối tia $CB$ lấy $N$ sao cho $CB=CN$

\(|\overrightarrow{MC}+\overrightarrow{BC}|=|\overrightarrow{MC}+\overrightarrow{CN}|=|\overrightarrow{MN}|\)

Xét tam giác $BMC$ và $ADI$ có:

$\widehat{B}=\widehat{A}=90^0$

$\widehat{D}=\widehat{M}$ (cùng bù $\widehat{AMC})$

Do đó 2 tam giác này đồng dạng

$\Rightarrow \frac{BM}{BC}=\frac{AD}{AI}$

$\Rightarrow BM=BC.\frac{AD}{AI}=\frac{2BC^2}{AB}=\frac{3\sqrt{2}a}{4}$

$BN=2BC=a\sqrt{3}$

Do đó, áp dụng định lý Pitago:

$|\overrightarrow{MN}|=MN=\sqrt{BM^2+BN^2}=\frac{\sqrt{66}a}{4}$

Hình vẽ:

đề như thế thì đương nhiên phải có điều kiện đó chứ em, đề đúng rồi anh xin xóa câu trl 

1. ĐKXĐ: \(a,b,c\) đôi một khác nhau.

\(\dfrac{\left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)

⇔\(\dfrac{x-c}{a-b}\left(\dfrac{x-b}{a-c}-\dfrac{x-a}{b-c}\right)=1\)

⇔\(\dfrac{x-c}{a-b}.\dfrac{\left(x-b\right)\left(b-c\right)-\left(x-a\right)\left(a-c\right)}{\left(a-c\right)\left(b-c\right)}=1\)

⇔\(\dfrac{x-c}{a-b}.\dfrac{bx-cx-b^2+bc-\left(ax-cx-a^2+ac\right)}{\left(a-c\right)\left(b-c\right)}=1\)

⇔\(\dfrac{x-c}{a-b}.\dfrac{bx-b^2+bc-ax+a^2-ac}{\left(a-c\right)\left(b-c\right)}=1\)

⇔\(\dfrac{x-c}{a-b}.\dfrac{x\left(b-a\right)+c\left(b-a\right)-\left(b-a\right)\left(a+b\right)}{\left(a-c\right)\left(b-c\right)}=1\)

⇔\(\dfrac{x-c}{a-b}.\dfrac{\left(b-a\right)\left(x-a-b+c\right)}{\left(a-c\right)\left(b-c\right)}=1\)

⇔\(\dfrac{\left(x-c\right)\left(a-b\right)\left(x-a-b+c\right)}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}-1=0\)

⇔\(\dfrac{\left(x-c\right)\left(a-b\right)\left(x-a-b+c\right)}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}-\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

⇔\(\left(x-c\right)\left(a-b\right)\left(x-a-b+c\right)-\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)

⇔\(\left(a-b\right)\left[\left(x-c\right)\left(x-a-b+c\right)-\left(b-c\right)\left(c-a\right)\right]=0\)

⇔\(a-b=0\) (loại do \(a\ne b\)) hay \(\left(x-c\right)\left(x-a-b+c\right)-\left(b-c\right)\left(c-a\right)=0\)

⇔\(x^2-ax-bx+cx-cx+ac+bc-c^2-\left(bc-ab-c^2+ac\right)=0\)

⇔\(x^2-ax-bx+cx-cx+ac+bc-c^2-bc+ab+c^2-ac=0\)

⇔\(x^2-ax-bx+ab=0\)

⇔\(x\left(x-a\right)-b\left(x-a\right)\)

⇔\(\left(x-a\right)\left(x-b\right)=0\)

⇔\(x=a\) hay \(x=b\)

-Vậy \(S=\left\{a;b\right\}\)

Bài 1:

a: ΔADC vuông tại D

=>\(DA^2+DC^2=AC^2\)

=>\(AC^2=6^2+8^2=36+64=100=10^2\)

=>AC=10(cm)

Xét ΔDAC vuông tại D có DH là đường cao

nên \(DH\cdot AC=DA\cdot DC\)

=>\(DH=\frac{6\cdot8}{10}=4,8\left(\operatorname{cm}\right)\)

Xét ΔDAC vuông tại D có sin ACD=\(\frac{DA}{AC}=\frac{6}{10}=\frac35\)

nên \(\hat{ACD}\) ≃37 độ

b: Xét ΔDAC vuông tại D có DH là đường cao

nên \(AD^2=AH\cdot AC;DC^2=CH\cdot CA\)

=>\(\frac{DA^2}{DC^2}=\frac{AH\cdot AC}{CH\cdot CA}=\frac{AH}{CH}\)

=>\(\left(\frac{BC}{BA}\right)^2=\frac{AH}{HC}\)

Bài 2: ĐKXĐ: x>=3

\(x^2+\sqrt{2x+1}+\sqrt{x-3}=5x\)

=>\(x^2-16+\sqrt{2x+1}-3+\sqrt{x-3}-1=5x-20\)

=>\(\left(x-4\right)\left(x+4\right)+\frac{2x+1-9}{\sqrt{2x+1}+3}+\frac{x-3-1}{\sqrt{x-3}+1}-5\left(x-4\right)=0\)

=>\(\left(x-4\right)\left\lbrack x+4+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}-5\right\rbrack=0\)

=>x-4=0

=>x=4(nhận)

Vì \(AB//CF\) ,áp dụng định lí Talet: 

\(\dfrac{AE}{EF}=\dfrac{BE}{EC}\Rightarrow\dfrac{AE}{AF}=\dfrac{BE}{BC}\Rightarrow\dfrac{AE^2}{AF^2}=\dfrac{BE^2}{BC^2}\\ \Rightarrow\dfrac{AE^2}{AF^2}=\dfrac{AE^2-AB^2}{BC^2}=\dfrac{AE^2}{BC^2}-\left(\dfrac{AB}{BC}\right)^2\left(pytago\right)\\ \Rightarrow\dfrac{AE^2}{AF^2}=\dfrac{AE^2}{BC^2}-9=\dfrac{AE^2}{\dfrac{1}{9}AB^2}-9\\ \Rightarrow\dfrac{AE^2}{AF^2}+9=\dfrac{9AE^2}{AB^2}\\ \Rightarrow\dfrac{1}{AF^2}+\dfrac{9}{AE^2}=\dfrac{9}{AB^2}\)

k mk đi

ai k mk

mk k lại

thanks

giải đi mình k

Lời giải:

$S_{ABC}=AB\times BC:2=30\times 16:2=240$ (cm²)

$S_{ADC}=AD\times DC:2=16\times 30:2=240$ (cm²)

\(\frac{S_{AMC}}{S_{ABC}}=\frac{AM}{AB}=\frac{2}{3}\)

\(\Rightarrow S_{AMC}=\frac{2}{3}\times S_{ABC}=\frac{2}{3}\times 240=160\) (cm²)

\(S_{AMCD}=S_{ADC}+S_{AMC}=240+160=400\) (cm²)