This - these

That - those

x¹⁰ = x

x¹⁰ - x = 0

x(x⁹ - 1) = 0

x = 0 hoặc x⁹ - 1 = 0

*) x⁹ - 1 = 0

x⁹ = 1

x = 1

Vậy x = 0; x = 1

\(x^{10}=x\)

\(\Rightarrow x=1\)

Bạn xem lại đề

B = \(\dfrac{x^2-2x+1}{x+1}\)

Với \(x\in\)Z, để B là số nguyên thì \(x^2-2x+1\)⋮ \(x+1\)

Theo Bezout ta có:   F(\(x\)) =  \(x^2\) - 2\(x\) + 1  ⋮ \(x+1\) ⇔ F(\(-1\))  ⋮ \(x+1\)

⇒ (-1)² - 2.(-1) + 1 ⋮ \(x\) + 1 ⇔ 4 ⋮ \(x\) + 1

⇔ \(x\) + 1 \(\in\) Ư(4) = { -4; -2; -1; 1; 2; 4}

\(\Leftrightarrow\) \(x\) \(\in\) { -5; -3; -2; 0; 1; 3}

192 = 2.96= 2. (8.12)

Vì: 12= 8+4

Vậy ao cũ có cạnh bằng 8m

Diện tích ao cũ:

8.8 = 64(m²)

 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 = \(\dfrac{1}{2}\)

\(\dfrac{99}{98}-\dfrac{98}{97}+\dfrac{1}{97\cdot98}\)

\(=\dfrac{99\cdot97}{98\cdot97}-\dfrac{98\cdot98}{97\cdot98}+\dfrac{1}{97\cdot98}\)

\(=\dfrac{99\cdot97-98^2+1}{98\cdot97}\)

\(=\dfrac{\left(98+1\right)\left(98-1\right)-98^2+1}{98\cdot97}\)

\(=\dfrac{98^2-1-98^2+1}{98\cdot97}\)

\(=\dfrac{0}{97\cdot98}\)

\(=0\)

\(\dfrac{99}{98}-\dfrac{98}{97}+\dfrac{1}{97.98}=\dfrac{99.97}{97.98}-\dfrac{98.98}{97.98}+\dfrac{1}{97.98}\)

\(=\dfrac{99.97-98.98+1}{97.98}=\dfrac{\left(98+1\right).\left(98-1\right)-98^2+1}{97.98}\)

\(=\dfrac{98^2-1-98^2+1}{97.98}=0\)

\(63+27.97+18=63+18+27.97=81++27.97=27.3+27.97=27.\left(3+97\right)=27.100=2700\)

2700

\(2^{10}+2^{11}+2^{12}=2^8\left(2^2+2^3+2^4\right)=2^8.\left(4+8+16\right)=2^8.28⋮28\left(đpcm\right)\)

\(2^{10}+2^{11}+2^{12}⋮28\\ \Rightarrow2^8+\left(4+8+16\right)\\ \Rightarrow2^8+28\\ \)

\(\)Vì \(28⋮28\\ \Rightarrow2^{10}+2^{11}+2^{12}⋮28.\)