|3x-5| (2y + 5) 2022(42-3)2020<=0
2^3.19-2^3.14+1^2020
10^2-[60:(5^6:5^4-3.5)]
160:{17+[3^2.5-(14=2^7:2^4)]}
798+100:[16-2(5^2022)]
t^2+5t-6 khi t=2
(a+b)^2-(b-a)^3+2012 khi a=5;b=a+1
x^3-3x^2y=3xy^2-y^3 khi x=3;y=2
23.19 - 23.14 + 12020
= 23.(19 - 14) + 1
= 8.5 + 1
= 41
102 - [60: (56: 54 - 3.5)]
= 100 - [60: (52 - 15)]
= 100 - [60: (25 - 15)]
= 100 - [60 : 10]
= 100 - 6
= 94
160: {17 + [32.5 - ( 14 + 27:24)]}
= 160: {17 + [9.5 - (14 + 23)]}
= 160: {17 + [45 - (14 + 8)]}
= 160: {17 + [45 - 22]}
= 160: {17 + 23}
= 160 : 40
= 4
1. 2019/2020-(2019/2020-2020/2021)
2.2/9+7/9 :(42/5-7/5
3.a)3/4+x/4=5/8
4./3x+1/-1/4=-1/4
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
Giải:
1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}\)
\(=\dfrac{2020}{2021}\)
2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\)
\(=\dfrac{2}{9}+\dfrac{7}{9}:7\)
\(=\dfrac{2}{9}+\dfrac{1}{9}\)
\(=\dfrac{1}{3}\)
3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\)
\(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\)
\(\dfrac{x}{4}=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\)
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x-1\right|=0\)
\(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
Chúc bạn học tốt!
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x+1\right|=0\)
\(3x+1=0\)
\(3x=0-1\)
\(3x=-1\)
\(x=-1:3\)
\(x=\dfrac{-1}{3}\)
(3x-2^4).7^2019=2.7^2020
5^x+1.5^2021=5^2022
giúp mình với mình tick cho
tìm gtnn của biểu thức P=x^3-3x+5 và Q=2x^2+y^2-2xy-6x+2y+2022
S = 5 + 5^2 + 5^3 + ... + 5^2020 + 5^2021. Chứng minh 4 . S + 5 + 5^2022
Chứng minh \(4\cdot S+5+5^{2022}\) là sao nhỉ?
TK :
Ta có A = 5 + 52 + 53 + ... + 52021
5A = 52 + 53 + 54 + ... + 52022
5A - A = ( 52 + 53 + 54 + ... + 52022 ) - ( 5 + 52 + 53 + ... + 52021 )
4A = 52022 - 5
Vậy 4A + 5 = 52022 - 5 + 5 = 52022
TK :
Ta có A = 5 + 52 + 53 + ... + 52021
5A = 52 + 53 + 54 + ... + 52022
5A - A = ( 52 + 53 + 54 + ... + 52022 ) - ( 5 + 52 + 53 + ... + 52021 )
4A = 52022 - 5
Vậy 4A + 5 = 52022 - 5 + 5 = 52022
tìm x,y,z,biết:\(|3x-5+(2y+5)^{2018}+\left(4z-3\right)^{2020}|\le0\)
Sửa đề: \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}\le0\)(1)
Ta có: \(\left|3x-5\right|\ge0;\left(2y+5\right)^{2018}\ge0;\left(4z-3\right)^{2020}\ge0.\)mọi x,y, z.
=> \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}\ge0\)với mọi x, y,z.
Như vậy (1) chỉ xảy ra trường hợp: \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}=0\)
<=> \(\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{2}\\z=\frac{3}{4}\end{cases}}\)
Vậy...
thầy mình cho đè kia cơ
Nếu đề đúng là như vậy thì làm như sau :
Bài giải
Vì : \(\left|3x-5+\left(2y+5\right)^{2018}+\left(4z-3\right)^{2020}\right|\ge0\)
\(\Rightarrow\) Chỉ xảy ra trường hợp :
\(\left(3x-5\right)+\left(2y+5\right)^{2018}+\left(4z-3\right)^{2020}=0\)
Mà \(\hept{\begin{cases}\left(2y+5\right)^{2018}\ge0\\\left(4z-3\right)^{2020}\ge0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3x-5=0\\\left(2y+5\right)^{2018}=0\\\left(4z-3\right)^{2020}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{2}\\z=\frac{3}{4}\end{cases}}\)
\(\Rightarrow\text{ }x=\frac{5}{3}\text{ ; }y=-\frac{5}{2}\text{ ; }z=\frac{3}{4}\)
s= 5 +5^2+5^3+...+5^2020+5^2021. Chứng tỏ rằng 4.S+5=5^2022
S = 5 + 52 + 53 +...+ 52020 + 52021
5S = 52+ 53 + 54 +...+ 52021 + 52022
5S-S =(52 + 53 + 54 + ... + 52021 + 52022)-(5 + 52 + 53 + ... + 52021)
4S = 52 + 53 + 54 +...+ 52021 + 52022 - 5 - 52 - 53 - ...- 52021
4S = (52 - 52)+(53- 53)+(54 - 54) + ... +(52021 - 52021)+(52022 - 5)
4S = 52022 - 5
4S + 5 = 52022 - 5 + 5
4S + 5 = 52022 (đpcm)
Tìm x, biết: a) 121-(115+x)= 3x-(25-9-5x)-8
b)2x+2.3x+1.5x = 10800
c) (3|x-1/2) . (8/15-1/5)+2/3-1
d) x+1/2022 + x+2/2021= x+3/2020 + x+4/2019
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)