Bạn nên viết hoàn chỉnh đề để mọi người hỗ trợ bạn tốt hơn.

Còn biểu thức $\frac{3}{x-2}+1$ nếu không có điều kiện gì thì sẽ không tồn tại giá trị lớn nhất.

\(\dfrac{7}{x-1}=\dfrac{x-1}{9}\)

\(\Rightarrow\left(x-1\right)\cdot\left(x-1\right)=7\cdot9\)

\(\Rightarrow\left(x-1\right)^2=63\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{63}\\x-1=-\sqrt{63}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\sqrt{7}+1\\x=-3\sqrt{7}+1\end{matrix}\right.\)

7/x-1 = x-1/9 ( ĐK : x khác 1 )

=> (x-1)^2 = 7.9 =63

=> x - 1 = ±\(\sqrt{63}\) 

=> x = 1 ±\(\sqrt{63}\) ( tmdk)

Vậy x = 1+√63 hoặc x = 1-√63

`#3107.101107`

\(\dfrac{27^2-9^2+3^3}{25}\\ =\dfrac{\left(3^3\right)^2-\left(3^2\right)^2+3^3}{25}\\ =\dfrac{3^3\cdot\left(3^3-3+1\right)}{25}\\ =\dfrac{3^3\cdot25}{25}\\ =3^3\\ =27\)

(27²-9²+3³)/25

=(729-81+27)/25

=675/25

=27

hê hê

Lời giải:
$C=1-2+2^2-2^3+2^4-....+2^{2022}$

$2C=2-2^2+2^3-2^4+2^5-...+2^{2023}$

$\Rightarrow C+2C=(1-2+2^2-2^3+2^4-....+2^{2022})+(2-2^2+2^3-2^4+2^5-...+2^{2023})$

$\Rightarrow 3C=2^{2023}-1$

$\Rightarrow C=\frac{2^{2023}-1}{3}$

\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)

\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)

\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)

\(x-\dfrac{2}{15}=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}+\dfrac{2}{15}\)

\(x=\dfrac{8}{15}\)

\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)

\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)

\(2x+5=4\)

\(2x=-1\)

\(x=-0,5\)

M = 2² + 4² + 6² + ... + 50²

= (1.2)² + (2.2)² + (2.3)² + ...+(2.25)²

= 2².1². + 2².2² + 2².3² + ... + 2².25²

= 2².(1² + 2² + 3² + ... + 25²)

= 4.25.(25 + 1).(2.25 + 1) : 6

= 22100

ta có:

2^2+4^2+6^2+....+50^2 

= 2^2 . 1^2 + 2^2 . 2^2 + 2^2 . 3^2 + ....+2^2.25^2

= 2^2(1^2+2^2+3^2+...+25^2)

= 4.5525

=22100

a²\(\equiv\)1 hoặc 0 (mod 12)
⇒a²-b^2\(\equiv\)1-1(mod 12) ( với mọi số chính phương)

mik đi mách thầy đức nhá