tính thuận tiện
2023 *10-2023*8-2023
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
94.2023+2023:1/6
=94.2023+2023.6
=(94+6).2023
=100.2023
=202300
2021 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
2021 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
a) 234 + 567 + 766 + 433
= (234 + 766) + (567 + 433)
= 1 000 + 1 000
= 2 000
b) 2 023 + 7 602 - 1 023 + 1 398
= (2 023 - 1 023) + (7 602 + 1 398)
= 1 000 + 9 000
= 10 000
c) 4 789 - 2 567 - 7 433 + 5 211
= (4 789 + 5 211) - (2 567 + 7 433)
= 10 000 - 10 000
= 0
c, 5 - 10 + 15 - 20 + 25 - 30 + 35 - 40 + 50 - 55 + 60 - 65 + 70
= 5 + (15 - 10) + (25 - 20) + (35 - 30) + (50 - 40) + (60 - 55) + (70 - 65)
= 5 + 5 + 5 + 5 + 10 + 5 + 5
= (5 + 5) + (5 + 5) + 10 + (5 + 5)
= 10 + 10 + 10 + 10
= 10 $\times$ 4
= 40
a, 234 + 567 + 766 + 433
= (234 + 766) + (567 + 433)
= 1000 + 1000
= 2000
b, 2023 + 7602 - 1023 + 1398
= (2023 - 1023) + (7602 + 1398)
= 1000 + 9000
= 10000
c, 4789 - 2567 - 7433 + 5211
= (4789 + 5211) - (2567 + 7433)
= 10000 - 10000
= 0
d, 5 - 10 + 15 - 20 + 25 - 30 + 35 - 40 + 50 - 55 + 60 - 65 + 70
= (70 -65) + (60 - 55) +(50 - 40) + (35 - 30) + (25 - 20) + (15 - 10) + 5
= 5 + 5 + 10 + 5 + 5 + 5 + 5
= 5 x 6 + 10
= 40
a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)
\(\left|x-1\right|^{2023}>=0\forall x\)
=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)
mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)
nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)
=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)
\(P=x^{2023}+\left(y-10\right)^{2023}\)
\(=1^{2023}+\left(9-10\right)^{2023}\)
=1-1
=0
c: \(\left|x-3\right|>=0\forall x\)
=>\(\left|x-3\right|+2>=2\forall x\)
=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)
mà \(\left|y+3\right|>=0\forall y\)
nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)
=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)
Dấu '=' xảy ra khi x-3=0 và y-3=0
=>x=3 và y=3
2023.(16 - 2024) + 2024.2023 - 16.(2023 + 10)
= 2023.16 - 2023.2024 + 2024.2023 - 16.2023 - 16.10
= (2023.16 - 16.2023) - (2023.2024 - 2024.2023) - 16.10
= 0 - 0 - 16.10
= - 160
\(2023.\left(16-2024\right)+2024.2023-16.\left(2023+10\right)\)
\(=2023.16-2023.2024+2023.2024-16.2023-16.10\)
\(=2023\left(16-16\right)+2023\left(2024-2024\right)-16.10\)
\(=0+0-160=-160\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2023}=\dfrac{1}{a+b+c}\)
\(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)
\(\left(a+b\right)\left[\dfrac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right]=0\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Đến đây bạn thay vào nữa là được nhé
2023×6+7×2023−2023
=2023×6+7×2023−2023x1
=2023×(6+7−1)
=2023×12
=24276
\(2023\times28+2023\times34-2023\times52\)
\(=2023\times\left(28+34-52\right)\)
\(=2023\times10\)
\(=20230\)
`# \text {DNamNgV}`
`2023 \times 28 + 2023 \times 34 - 2023 \times 52`
`= 2023 \times (28 + 34 - 52)`
`= 2023 \times 10 `
`=20230`
\(\text{ 2023 x 10 - 2023 x 8 - 2023}\)
\(\text{= 2023 x (10-8-1)}\)
\(\text{= 2023 x 1}\)
\(\text{= 2028}\)
2023 \(\times\) 10 - 2023 \(\times\) 8 - 2023
= 2023 \(\times\) 10 - 2023 \(\times\) 8 - 2023 \(\times\) 1
= 2023 \(\times\) (10 - 8 - 1)
= 2023 x 1
= 2023