(x mũ 2023+3 mũ x 2022+1) mũ 2000 biết x=-3
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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.
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:
$\frac{m}{\frac{3}{5}}=\frac{n}{\frac{5}{1}}=\frac{p}{\frac{5}{2}}=\frac{m+n+p}{\frac{3}{5}+\frac{5}{1}+\frac{5}{2}}=\frac{\frac{213}{70}}{\frac{81}{10}}=\frac{71}{189}$
$\Rightarrow m=\frac{71}{315}; n=\frac{355}{189}; p=\frac{355}{378}$
Khi x=-3 thì \(\left(x^{2023}+3x^{2022}+1\right)^{2000}=\left[\left(-3\right)^{2023}+3\cdot\left(-3\right)^{2022}+1\right]^{2000}\)
\(=\left[-3^{2023}+3^{2023}+1\right]^{2000}\)
\(=1^{2000}=1\)