(x mũ 2023+3 mũ x 2022+1) mũ 2000 biết x=-3
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Lời giải:
$\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-a)(b-c)}+\frac{a-b}{(c-a)(c-b)}=2024$
$\Rightarrow \frac{(a-c)-(a-b)}{(a-b)(a-c)}+\frac{(b-a)-(b-c)}{(b-a)(b-c)}+\frac{(c-b)-(c-a)}{(c-a)(c-b)}=2024$
$\Rightarrow \frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}=2024$
$\Rightarrow \frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}=2024$
$\Rightarrow 2(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a})=2024$
$\Rightarrow 2Q=2024$
$\Rightarrow Q=1012$
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Lời giải:
Hiệu chiều dài và chiều rộng: $3,5+3,5=7$ (dm)
Tổng chiều dài và chiều rộng: $82:2=41$ (dm)
Chiều dài hcn: $(41+7):2=24$ (dm)
Chiều rộng hcn: $24-7=17$ (dm)
Diện tích hcn: $24\times 17=408$ (dm2)
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9/10 của \(-1\dfrac{7}{18}=-\dfrac{25}{18}\) là:
\(-\dfrac{25}{18}\cdot\dfrac{9}{10}=\dfrac{-225}{180}=-\dfrac{5}{4}\)
8/35 của 31,5 là \(31,5\cdot\dfrac{8}{35}=0,9\cdot8=7,2\)
\(-\dfrac{5}{4}< X< 7,2\)
=>-1,25<X<7,2
mà X nguyên
nên \(X\in\left\{-1;0;1;2;3;4;5;6;7\right\}\)
=>A={-1;0;1;2;3;4;5;6;7}
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1: \(-\dfrac{1}{3}\cdot\dfrac{5}{7}=\dfrac{-1\cdot5}{3\cdot7}=\dfrac{-5}{21}\)
2: \(-\dfrac{2}{5}\cdot\dfrac{5}{4}=\dfrac{-2}{4}\cdot\dfrac{5}{5}=-\dfrac{2}{4}=-\dfrac{1}{2}\)
3: \(-\dfrac{2}{5}\cdot\dfrac{3}{7}=\dfrac{-2\cdot3}{5\cdot7}=\dfrac{-6}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2}{3}x+\dfrac{2023}{2024}=\dfrac{5}{3}-\dfrac{1}{2024}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{3}-\dfrac{1}{2024}-\dfrac{2023}{2024}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
=>x=1
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\(\dfrac{-15}{26}\cdot\dfrac{4}{15}+\dfrac{15}{26}\cdot\dfrac{-3}{15}-\dfrac{7}{15}\cdot\dfrac{-11}{26}\)
\(=\dfrac{15}{26}\left(-\dfrac{4}{15}-\dfrac{3}{15}\right)+\dfrac{11}{26}\cdot\dfrac{7}{15}\)
\(=-\dfrac{15}{26}\cdot\dfrac{7}{15}+\dfrac{11}{26}\cdot\dfrac{7}{15}\)
\(=\dfrac{7}{15}\left(-\dfrac{15}{26}+\dfrac{11}{26}\right)=\dfrac{7}{15}\cdot\dfrac{-4}{26}=\dfrac{-28}{390}=\dfrac{-14}{195}\)
Khi x=-3 thì \(\left(x^{2023}+3x^{2022}+1\right)^{2000}=\left[\left(-3\right)^{2023}+3\cdot\left(-3\right)^{2022}+1\right]^{2000}\)
\(=\left[-3^{2023}+3^{2023}+1\right]^{2000}\)
\(=1^{2000}=1\)