hảo hán nào giải đc không vậy?

quên cách làm rùi

ủa cái j dợ

a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)

\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)

b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)

\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)

\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)

=-3-1+1

=-3

c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)

\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)

 2/2023 / 4/3 ?

A = \(\dfrac{\dfrac{2}{3}-\dfrac{2}{13}+\dfrac{2}{2023}}{\dfrac{4}{3}-\dfrac{4}{13}+\dfrac{4}{2023}}\)

A = \(\dfrac{2.\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{2023}\right)}{4.\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{2023}\right)}\)

A = \(\dfrac{1}{2}\)

\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+\dfrac{15}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{2023^2}\)

\(A=(1+1+1+...+1)-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{2023^2})\)

Tổng số hạng của 2 ngoặc trên bằng nhau và =(2023-2):1+1=2022(số hạng)

\(A=2022-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2})\)

Ta thấy:

\(0<\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)

Ta có

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{2022}-\dfrac{1}{2023}\)

\(=1-\dfrac{1}{2023}<1\)

Do đó,2021<A<2022 

Vậy giá trị của A không phải 1 số tự nhiên(đpcm)

Ko bt

\(A=2+2^2+2^3+...2^{2023}\)

\(\Rightarrow A+1=1+2+2^2+2^3+...2^{2023}\)

\(\Rightarrow A+1=\dfrac{2^{2023+1}-1}{2-1}\)

\(\Rightarrow A+1=2^{2024}-1\)

\(\Rightarrow A=2^{2024}-2\)

\(\Rightarrow A=2^{2020}.2^4-2\)

\(\Rightarrow A=\left(2^{20}\right)^{101}.2^4-2\)

Ta thấy :

\(\left(2^{20}\right)^{101}\) có tận cùng là chữ số \(76\)

\(2^4=16\) có tận cùng là chữ số \(6\)

\(\Rightarrow\left(2^{20}\right)^{101}.2^4\) có tận cùng là chữ số \(6\)

\(\Rightarrow A=\left(2^{20}\right)^{101}.2^4-2\) có tận cùng là chữ số 4  \(\left(6-2=4\right)\)

$A=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+....+\frac{{2023}^2-1}{{2023}^2}$

$A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+....+1-\frac{1}{{2023}^2}$

$A=2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$

$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}$

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{2023}$

$\Rightarrow 0<\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<1-\frac{1}{2023}<1$

$\Rightarrow 2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$ko phải số tự nhiên

$\Rightarrow A$ ko phải số tự nhiên

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k

a)  A=x^2+4x+4=(x+2)^2.

Giờ ta tính giá trị của đa thức A với x=98:

A=(98+2)^2=100^2=10000

b)  B=x^3+9x^2+27x+27=(x+3)^3.

Thế x=-103 => (-103+3)^3=-1000000

c) Tách C = a⋅b−a⋅c+2⋅c−2⋅b rồi kết hợp lại thành C=(a−2)⋅b+(2−a)⋅c.

Thế a,b,c vào được vậy 

C=(2−2)⋅1.007+(2−2)⋅(−0.006) =0

d) Bài này khó quá mà tui nghĩ là đưa mấy cặp (2023^2-2022^2) thành dạng a^2-b^2=(a-b)(a+b) á

d: D=(2023^2-2022^2)+(2021^2-2020^2)+...+(3^2-2^2)+(1^2-0^2)

=2023+2022+...+3+2+1+0

=2023*2024/2=2047276

\(H=\left(3x-6\right)^2-3\left|2x-4\right|+2023\)

\(=\left(3x-6\right)^2-2\left|3x-6\right|+2023\)

\(=\left(3x-6\right)^2-2\left|3x-6\right|+1+2022\)

\(=\left(\left|3x-6\right|-1\right)^2+2022\)

Do \(\left(\left|3x-6\right|-1\right)^2\ge0;\forall x\)

\(\Rightarrow H\ge2022\)

\(\Rightarrow H_{min}=2022\) khi \(\left|3x-6\right|-1=0\Rightarrow x=\left\{\dfrac{7}{3};\dfrac{5}{3}\right\}\)