Đề thiếu. Bạn xem lại đề.

1) \(\dfrac{2}{3}\left(x-\dfrac{5}{6}\right)-\dfrac{1}{5}\left(\dfrac{3}{4}-\dfrac{x}{2}\right)=1\)

\(\Rightarrow\dfrac{2}{3}x-\dfrac{5}{9}-\dfrac{3}{20}+\dfrac{x}{10}=1\)

\(\Rightarrow x\left(\dfrac{2}{3}+\dfrac{1}{10}\right)-\dfrac{127}{180}=1\)

\(\Rightarrow x\cdot\dfrac{23}{30}=1+\dfrac{127}{180}\)

\(\Rightarrow x\cdot\dfrac{23}{30}=\dfrac{307}{180}\)

\(\Rightarrow x=\dfrac{307}{180}:\dfrac{23}{30}\)

\(\Rightarrow x=\dfrac{307}{138}\)

2) \(\left(\left|x\right|-\dfrac{1}{3}\right)\left(\left|x\right|+2\right)=0\)

TH1: \(\left|x\right|-\dfrac{1}{3}=0\)

\(\Rightarrow\left|x\right|=\dfrac{1}{3}\)

\(\Rightarrow x=\pm\dfrac{1}{3}\) 

TH2: \(\left|x\right|+2=0\)

\(\Rightarrow\left|x\right|=-2\) (vô lý) 

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