1. Tính giá trị biểu thức:

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

1. Tính giá trị biểu thức:

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

1. Tìm x:

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

1. Tìm x:

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$

$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$

$\Rightarrow 5a-a=5^{2024}-1$

$\Rightarrow 4a=5^{2024}-1$

$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)

Sửa đề: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

Ta có: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(3a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}\)

=>\(3a+a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

Đặt \(b=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(3b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}\)

=>\(3b+b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(4b=-1-\frac{1}{3^{2023}}=\frac{-3^{2023}-1}{3^{2023}}\)

=>\(b=\frac{-3^{2023}-1}{4\cdot3^{2023}}\)

Ta có: \(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1+\frac{-3^{2023}-1}{4\cdot3^{2023}}-\frac{2024}{3^{2024}}=1+\frac{-3^{2024}-3}{4\cdot3^{2024}}-\frac{8096}{4\cdot3^{2024}}\)

=>\(4a=1-\frac{3^{2024}+8099}{4\cdot3^{2024}}=1-\frac14-\frac{8099}{4\cdot3^{2024}}=\frac34-\frac{8099}{4\cdot3^{2024}}\)

=>\(4a<\frac34\)

=>\(a<\frac{3}{16}\)

mà \(\frac{3}{16}<1<\frac{20}{3}\)

nên \(a<\frac{20}{3}\)

A=5(1+5^2)+5^5(1+5^2)+...+5^2021(1+5^2)

=26(5+5^5+...+5^2021) chia hết cho 26

Đay là của lp 6 ư, nhìn ko hỉu j cả

cái nì mik chịu

M=(1/5+1/5^2+1/5^3+...+1/5^2023) + 1/5x(1/5+1/5^2+1/5^3+...+1/5^2022) + ... + 1/5^2021x(1/5+1/5^2) + 1/5^2022x1/5

Xét biểu thức N=1/5+1/5^2+1/5^3 + ... + 1/5^k (K>0, k thuộc Z)

=> 5N=1+1/5+1/5^2+1/5^3+...+1/5^(k-1)

=> 4N= 5N - N =1 - 1/5^k

=> 1/5+1/5^2+1/5^3 + ... + 1/5^k = 1/4x(1-1/5^k)

Thay vào biểu thức M, ta có:

M= 1/4x(1-1/5^2023) + 1/5x1/4x(1-1/5^2022) + ... + 1/5^2021x1/4x(1-1/5^2) + 1/5^2022x1/4x(1-1/5)

=> 4M = (1+1/5+1/5^2+...+1/5^2022) - 2023/5^2023

=> 4M = 5/4x(1-1/5^2023)-2023/5^2023 < 5/4

=> M < 5/16 < 1/3 

Vậy M < 1/3 [ vượt chỉ tiêu nhé =)) ]

SOS

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

A=(-1)+(-1)+...+(-1)+2023

=2023-1011

=1012

??????

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)