Lời giải:
$\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-a)(b-c)}+\frac{a-b}{(c-a)(c-b)}=2024$

$\Rightarrow \frac{(a-c)-(a-b)}{(a-b)(a-c)}+\frac{(b-a)-(b-c)}{(b-a)(b-c)}+\frac{(c-b)-(c-a)}{(c-a)(c-b)}=2024$

$\Rightarrow \frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}=2024$

$\Rightarrow \frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}=2024$

$\Rightarrow 2(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a})=2024$

$\Rightarrow 2Q=2024$

$\Rightarrow Q=1012$

Lời giải:
$\widehat{ABC}=180^0-\widehat{BAC}-\widehat{ACB}=180^0-90^0-30^0=60^0$

$\widehat{ABE}=\widehat{ABC}:2=60^0:2=30^0$

Đáp án B.

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:

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