Khi x=-3 thì \(\left(x^{2023}+3x^{2022}+1\right)^{2000}=\left[\left(-3\right)^{2023}+3\cdot\left(-3\right)^{2022}+1\right]^{2000}\)

\(=\left[-3^{2023}+3^{2023}+1\right]^{2000}\)

\(=1^{2000}=1\)

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)