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

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

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 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

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

Bài 10:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)

a:

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)

\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)

b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)

 \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)

\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)

\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)

Bài 10:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)

a:

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)

\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)

b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)

 \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)

\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)

\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)

Bài 14:

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

\(A=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\cdot\left(-x\right)\cdot\left(-y\right)\)

=-xyz

=-2

 Tổng 2 số đó là: 

     404 x 2 = 808

 Nếu viết thêm chữ số 7 vào bên trái số thứ nhất \(\rightarrow\) số đó tăng thêm 700 đơn vị \(\rightarrow\) số thứ nhất bé hơn số thứ hai 700 đơn vị

 Số thứ nhất là: ( 808 - 700 ) : 2 = 54

 Số thứ hai là: 700 + 54 = 754

         Đáp số: Số thứ nhất: 54

                     Số thứ hai: 754

a: Sau buổi sáng thì số mét vải còn lại chiếm:

\(1-\dfrac{3}{4}=\dfrac{1}{4}\)(tấm vải)

15m vải còn lại chiếm: \(\dfrac{1}{4}\times\left(1-\dfrac{1}{5}\right)=\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)(tấm vải)

Độ dài tấm vải là \(15:\dfrac{1}{5}=75\left(m\right)\)

b: Buổi sáng bán được: \(75\times\dfrac{3}{4}=56,25\left(m\right)\)

Buổi chiều bán được:

75-56,25-15=3,75(m)

a: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)

mà \(\widehat{bMB}=50^0\)

nên \(\widehat{NMC}=50^0\)

Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)

=>\(\widehat{MNC}+110^0=180^0\)

=>\(\widehat{MNC}=70^0\)

Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)

=>\(\widehat{C}+50^0+70^0=180^0\)

=>\(\widehat{C}=60^0\)

b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)

=>\(\widehat{NMB}+50^0=180^0\)

=>\(\widehat{NMB}=130^0\)

Ta có: MN//AB

=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)

=>\(\widehat{CBA}=50^0\)

BN là phân giác của góc CBA

=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)

=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)

c: BN là phân giác của góc CBA

=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAN}+60^0+50^0=180^0\)

=>\(\widehat{BAN}=70^0\)

Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)

=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)

a: Ta có: \(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

\(\widehat{ICB}+\widehat{NCB}=\widehat{NCI}=90^0\)

Do đó: \(\widehat{ICA}=\widehat{NCB}\)

Ta có: \(\widehat{CAI}+\widehat{CBI}=90^0\)(ΔCBA vuông tại C)

\(\widehat{CBI}+\widehat{CBN}=\widehat{NBI}=90^0\)

Do đó: \(\widehat{CAI}=\widehat{CBN}\)

Xét ΔCAI và ΔCBN có

\(\widehat{CAI}=\widehat{CBN}\)

\(\widehat{ICA}=\widehat{NCB}\)

Do đó: ΔCAI~ΔCBN

b: Ta có: \(\widehat{ACM}+\widehat{ACI}=\widehat{ICM}=90^0\)

\(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

Do đó: \(\widehat{ACM}=\widehat{ICB}\)

Ta có: \(\widehat{CAM}+\widehat{CAB}=\widehat{BAM}=90^0\)

\(\widehat{CAB}+\widehat{CBA}=90^0\)(ΔCAB vuông tại C)

Do đó: \(\widehat{CAM}=\widehat{CBA}\)

Xét ΔCAM và ΔCBI có

\(\widehat{CAM}=\widehat{CBI}\)

\(\widehat{ACM}=\widehat{BCI}\)

Do đó: ΔCAM~ΔCBI

=>\(\dfrac{AC}{CB}=\dfrac{AM}{BI}\)

=>\(AC\cdot BI=MA\cdot BC\)

c: Xét tứ giác CIBN có \(\widehat{ICN}+\widehat{IBN}=90^0+90^0=180^0\)

nên CIBN là tứ giác nội tiếp

=>\(\widehat{CIN}=\widehat{CBN}\)

=>\(\widehat{CIN}=\widehat{BAC}\)