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bài 7:a thực hiện phép tính .81 x 2022 + 25 x 2022 - 6 x 2022 .B Tìm x biết ( x - 1 ) 2/3 - 1/5= 2/5
\(a,81\cdot2022+25\cdot2022-6\cdot2022=2022\cdot\left(81+25-6\right)=2022\cdot100=202200\)
\(b,\left(x-1\right)\cdot\frac{2}{3}-\frac{1}{5}=\frac{2}{5}\)
\(\left(x-1\right)\cdot\frac{2}{3}=\frac{3}{5}\)
\(x-1=\frac{9}{10}\)
\(x=\frac{19}{10}\)
Vậy \(x=\frac{19}{10}\)
( Nếu phần b là hỗn số thì mình làm thế kia , còn nếu là nhân thì bạn tham khảo Câu hỏi của lương bảo ngọc - Toán lớp 5 - Học trực tuyến OLM nhé )
81 x 2022 + 25 x 2022 - 6 x 2022
= ( 81 + 25 - 6 ) x 2022
= 100 x 2022
= 202 200
b) \(\left(\text{x - 1}\right)\frac{\text{2}}{\text{3}}-\frac{\text{1}}{\text{5}}=\frac{\text{2}}{\text{5}}\)
\(\frac{\text{3 x }\text{( x - 1 ) }+\text{2}}{\text{3}}=\frac{\text{1}}{\text{5}}+\frac{\text{2}}{\text{5}}=\frac{\text{3}}{\text{5}}\)
=> \(\text{3 x ( x - 1 ) }+\text{2}=\frac{\text{3}}{\text{5}}\text{ x 3 = }\frac{\text{9}}{\text{5}}\)
=> \(\text{3 x ( x - 1 ) }=\frac{\text{9}}{\text{5}}-\text{2}=\frac{\text{-1}}{\text{5}}\)
=> \(\text{ x-1}=\frac{\text{-1}}{\text{5}}:3=\frac{\text{-1}}{\text{15}}\)
=> \(\text{x}=\frac{\text{-1}}{\text{15}}+\text{1 = }\frac{\text{14}}{\text{15}}\)
`2022 \times 123 \times 12 - 2022 \times 11`
`= 2022 \times (123 \times 12 - 11)`
`= 2022 \times (1476 - 11)`
`= 2022 \times 1465`
`= 2962230`
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
a. \(\dfrac{2021+2020.2022}{2021.2022-1}\)
\(\dfrac{2021.2022-2022+2021}{2021.2022-1}=\dfrac{2021.2022-1}{2021.2022-1}=1\)
\(b.\dfrac{2022+2021.2023}{2022.2023-1}=\dfrac{2021.2023-2023+2022}{2022.2023-1}\)
\(=\dfrac{2021.2023-1}{2022.2023-1}\)
a) (x - 15) × 7 - 270 : 45 = 169
(x - 15) × 7 - 6 = 169
(x - 15) × 7 = 169 + 6
(x - 15) × 7 = 175
x - 15 = 175 : 7
x - 15 = 25
x = 25 + 15
x = 40
b) [(4x + 28) × 3 + 55] : 5 = 35
(4x + 28) × 3 + 55 = 35 × 5
(4x + 28) × 3 + 55 = 175
(4x + 28) × 3 = 175 - 55
(4x + 28) × 3 = 120
4x + 28 = 120 : 3
4x + 28 = 40
4x = 40 - 28
4x = 12
x = 12 : 4
x = 3
c) (455 × x : 2 × 6) : 5 = 31
455 × x : 2 × 6 = 31 × 5
455 × x : 2 × 6 = 155
x × 455 : 2 × 6 = 155
x × 1365 = 155
x = 155 : 1365
x = 31/273
d) 128 × x - 12 × x - 16 × x = 520800
(128 - 12 - 16) × x = 520800
100 × x = 520800
x = 520800 : 100
x = 5208
e) (x × 0,25 + 2022) × 2023 = (50 + 2022) × 2023
(x × 0,25 + 2022) × 2023 = 2072 × 2023
(x × 0,25 + 2022) × 2023 = 4191656
x × 0,25 + 2022 = 4191656 : 2023
x × 0,25 + 2022 = 2072
x × 0,25 = 2072 - 2022
x × 0,25 = 50
x = 50 : 0,25
x = 200
f) 4 × x + 100 = x + 280
4 × x - x = 280 - 100
(4 - 1) × x = 180
3 × x = 180
x = 180 : 3
x = 60
g) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 7450
x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450
100 × x + 100 × 101 : 2 = 7450
100 × x + 5050 = 7450
100 × x = 7450 - 5050
100 × x = 2400
x = 2400 : 100
x = 24
2:
b=2000*2004
=(2002-2)*(2002+2)
=2002^2-4
=>b<a
1:
a: \(=8\cdot9\left(14+17+19\right)=72\cdot50=3600\)
Bài 1:
\(8\times9\times14+6\times17\times12+19\times4\times18\)
\(=8\times9\times14+3\times2\times17\times2\times2\times3+19\times4\times2\times9\)
\(=8\times9\times14+17\times8\times9+19\times8\times9\)
\(=8\times9\times\left(14+17+19\right)\)
\(=8\times9\times50\)
\(=72\times5\times10\)
\(=360\times10\)
\(=3600\)
Bài 2:
Ta có:
\(a=2022\times2022\)
Và: \(b=2000\times2004\)
Mà: \(2022>2000,2022>2004\)
\(\Rightarrow2022\times2022>2000\times2004\)
\(\Rightarrow a>b\)
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
=125 x 12 +12 x 874 +12 x 1
=12x( 125 + 874 +1)
=12 x 1000
=12000
\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)
\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)
a,50%+127−21=21+127−21=(21−21)+127=127b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12
=25×5−25:3×12=25×(5−13)×12=25×143×12=1400=25×5−25:3×12=25×(5−31)×12=25×314×12=1400