Bài 8:

a: \(\dfrac{x}{5}=\dfrac{y}{6}\)

=>\(\dfrac{x}{20}=\dfrac{y}{24}\)

\(\dfrac{y}{8}=\dfrac{z}{7}\)

=>\(\dfrac{y}{24}=\dfrac{z}{21}\)

Do đó: \(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=k\)

=>x=20k;y=24k;z=21k

x+y-z=69

=>20k+24k-21k=69

=>23k=69

=>k=3

=>\(x=20\cdot3=60;y=24\cdot3=72;z=21\cdot3=63\)

b: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

=>x=3k;y=4k;z=5k

\(2x^2+2y^2-3z^2=-100\)

=>\(2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

=>\(k^2=4\)

=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1: k=2

=>\(x=3\cdot2=6;y=4\cdot2=8;z=5\cdot2=10\)

TH2: k=-2

=>\(x=3\cdot\left(-2\right)=-6;y=4\cdot\left(-2\right)=-8;z=5\cdot\left(-2\right)=-10\)

a, Theo tc dãy tỉ số bằng nhau 

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{22}{-2}=-11\Rightarrow x=-33;y=-55\)

b, \(\dfrac{5}{2}=\dfrac{y}{x}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}\)Theo tc dãy tỉ số bằng nhau 

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\Rightarrow x=-6;y=-15\)

c, \(7x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{7}\)Theo tc dãy tỉ số bằng nhau 

\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=-\dfrac{21}{-4}=7\Rightarrow x=28;y=49\)

Bài 11: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)

=>\(\left(2a+13b\right)\left(3c-7d\right)=\left(3a-7b\right)\left(2c+13d\right)\)

=>\(6ac-14ad+39bc-91bd=6ac+39ad-14bc-91bd\)

=>-14ad-39ad=-14bc-39bc

=>ad=bc

=>\(\dfrac{a}{b}=\dfrac{c}{d}\)

Bài 12:

\(\dfrac{a+2019}{a-2019}=\dfrac{b+2020}{b-2020}\)

=>\(\left(a+2019\right)\left(b-2020\right)=\left(a-2019\right)\left(b+2020\right)\)

=>\(ab-2020a+2019b-2019\cdot2020=ab+2020a-2019b-2019\cdot2020\)

=>-2020a-2020a=-2019b-2019b

=>2020a=2019b

=>\(\dfrac{a}{2019}=\dfrac{b}{2020}\)

\(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(x-4\right)-\left(x^2-16\right)\left(x+4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(x-4-x-4+3\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

Xét 2 ΔABO và ΔADO ta có:

\(\widehat{BAO}=\widehat{DAO}\) (AD là phân giác của góc BAC) 

\(OA\) chung

\(\widehat{AOB}=\widehat{AOD}\left(gt\right)\) 

\(=>\Delta ABO=\Delta ADO\left(g.c.g\right)\) 

\(=>\widehat{B}=\widehat{D_1}\) (hai góc tương ứng) 

ΔABC cân tại A

=>\(\widehat{BAC}=180^0-2\cdot\widehat{ABC}=100^0\)

AD là phân giác góc ngoài tại đỉnh A

=>\(\widehat{CAD}=\dfrac{180^0-\widehat{BAC}}{2}=40^0\)

=>\(\widehat{CAD}=\widehat{ACB}\left(=40^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên AD//BC

Xét ΔABC vuông tại A có \(\widehat{ABC}+\widehat{ACB}=90^0\)

=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)

=>\(\widehat{OBC}+\widehat{OCB}=45^0\)

Xét ΔOBC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)

=>\(\widehat{BOC}+45^0=180^0\)

=>\(\widehat{BOC}=135^0\)

\(\widehat{A}=180^o-\widehat{B}-\widehat{C}=180^o-40^o-40^o=100^o\)

=> \(\widehat{A_{ngoai}}=180^o-100^o=80^o\) 

=> \(\widehat{DAB}=\dfrac{1}{2}\widehat{A_{ngoai}}=\dfrac{1}{2}\cdot80^o=40^o\)

Ta có: \(\widehat{DAB}=\widehat{ABC}\left(=40^o\right)\)

Mà 2 góc này ở vị trí so le trong

=> AD//BC 

cậu giúp mik nhiều ghê, cám ơn nha

\(\left(2,5x-\dfrac{4}{7}\right):\dfrac{8}{21}=-1,5\)

=>\(\left(\dfrac{5}{2}x-\dfrac{4}{7}\right):\dfrac{8}{21}=-\dfrac{3}{2}\)

=>\(\dfrac{5}{2}x-\dfrac{4}{7}=\dfrac{-3}{2}\cdot\dfrac{8}{21}=\dfrac{-24}{42}=\dfrac{-4}{7}\)

=>\(\dfrac{5}{2}x=-\dfrac{4}{7}+\dfrac{4}{7}=0\)

=>x=0

\(2^2+3^2+...+2021^2\)

\(=\left(1^2+2^2+...+2021^2\right)-1\)

\(=\dfrac{2021\cdot\left(2021+1\right)\left(2\cdot2021+1\right)}{6}=1\)

\(=2753594310\)