Lời giải:

Gọi 3 phân số đó là $m=\frac{a}{b}, n=\frac{c}{d}, p=\frac{e}{f}$. 

Theo bài ra ta có:
$m+n+p=\frac{213}{70}$ (1)

$\frac{a}{3}=\frac{c}{5}=\frac{e}{5}$

$\frac{b}{5}=\frac{d}{1}=\frac{f}{2}$

$\Rightarrow \frac{a}{3}: \frac{b}{5}=\frac{c}{5}: \frac{d}{1}=\frac{e}{5}: \frac{f}{2}$

$\Rightarrow \frac{a}{b}: \frac{3}{5}=\frac{c}{d}:\frac{5}{1}=\frac{e}{f}: \frac{5}{2}$

$\Rightarrow m: \frac{3}{5}=n: \frac{5}{1}=p:\frac{5}{2}$ (2)

Từ $(1); (2)$, áp dụng TCDTSBN:

$\frac{m}{\frac{3}{5}}=\frac{n}{\frac{5}{1}}=\frac{p}{\frac{5}{2}}=\frac{m+n+p}{\frac{3}{5}+\frac{5}{1}+\frac{5}{2}}=\frac{\frac{213}{70}}{\frac{81}{10}}=\frac{71}{189}$

$\Rightarrow m=\frac{71}{315}; n=\frac{355}{189}; p=\frac{355}{378}$

Lời giải:

$4x=3y\Rightarrow \frac{x}{3}=\frac{y}{4}; \frac{y}{5}=\frac{z}{6}$

$\Rightarrow \frac{x}{15}=\frac{y}{20}=\frac{z}{24}$

Đặt $\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k$

$\Rightarrow x=15k; y=20k; z=24k$

Khi đó:

$M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{186k}{245k}=\frac{186}{245}$

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=k\)

=>\(x=3k;y=5k;z=7k\)

\(x^2-y^2+z^2=-60\)

=>\(\left(3k\right)^2-\left(5k\right)^2+\left(7k\right)^2=-60\)

=>\(9k^2-25k^2+49k^2=-60\)

=>\(33k^2=-60\)

=>\(k^2=-\dfrac{60}{33}\left(vôlý\right)\)

=>\(\left(x,y,z\right)\in\varnothing\)

\(2x=\dfrac{y}{3}=\dfrac{z}{5}\)

=>\(\dfrac{x}{0,5}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà \(\dfrac{x+y-z}{2}=-20\)

nên \(\dfrac{x}{0,5}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{0,5+3-5}=\dfrac{-40}{-1,5}=\dfrac{40}{1,5}\)

=>\(x=\dfrac{20}{1,5}=\dfrac{40}{3};y=\dfrac{40}{1,5}\cdot3=80;z=40\cdot\dfrac{5}{1,5}=40\cdot\dfrac{10}{3}=\dfrac{400}{3}\)

Ta có: \(\widehat{ABE}=\dfrac{\widehat{ABC}}{2}\)

\(\widehat{ACF}=\dfrac{\widehat{ACB}}{2}\)

mà \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

nên \(\widehat{ABE}=\widehat{ACF}\)

Xét ΔABE và ΔACF có

\(\widehat{ABE}=\widehat{ACF}\)

AB=AC

\(\widehat{BAE}\) chung

Do đó: ΔABE=ΔACF

=>BE=CF

a: Xét ΔABI và ΔACI có

AB=AC

BI=CI

AI chung

Do đó: ΔABI=ΔACI

b: Sửa đề: Chứng minh AC=AE

Ta có: CE//AI

=>\(\widehat{AEC}=\widehat{BAI};\widehat{CAI}=\widehat{ACE}\)

mà \(\widehat{BAI}=\widehat{CAI}\)(ΔABI=ΔACI)

nên \(\widehat{AEC}=\widehat{ACE}\)

=>AC=AE

Số tiền để mua 20 cái kẹo là

\(2000\times20=40000\)( đồng )

Giá mỗi cái bánh là

\(40000\div8=5000\)( đồng )

Đáp số 5000 đồng

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

\(=xy^2-x+x^2y-y\)

\(=xy\left(x+y\right)-\left(x+y\right)\)

=5xy-5

\(f\left(x\right)=ax^2+bx+c\)

Mà: \(f\left(0\right)=2\) thay `x=0` ta có:

\(\Rightarrow f\left(0\right)=a\cdot0^2+b\cdot0+c=2\Rightarrow c=2\) 

       \(f\left(1\right)=7\) thay `x=1` ta có:

\(\Rightarrow f\left(1\right)=a\cdot1^2+b\cdot1+c=7\Rightarrow a+b+c=7\Rightarrow a+b=5\) (vì `c = 2`) 

\(\Rightarrow a=5-b\) (*) 

      \(f\left(-2\right)=-14\) 

\(\Rightarrow f\left(-2\right)=a\cdot\left(-2\right)^2+b\cdot-2+c=-14\)

\(\Rightarrow4a-2b+c=-14\)

\(\Rightarrow4a-2b=-16\) (vì `c=2`) 

\(\Rightarrow2a-b=-8\) (**) 

Thay (*) vào (**) ta có:

\(2\cdot\left(5-b\right)-b=-8\)

\(\Rightarrow10-2b-b=-8\)

\(\Rightarrow-3b=-18\)

\(\Rightarrow b=6\)

\(\Rightarrow a=5-6=-1\)

Vậy: ...