Câu III:
a) Ta có: \(A=n^4-14n^3+71n^2-154n+120\)
\(=n^4-2n^3-12n^3+24n^2+47n^2-94n-60n+120\)
\(=n^3\left(n-2\right)-12n^2\left(n-2\right)+47n\left(n-2\right)-60\left(n-2\right)\)
\(=\left(n-2\right)\left(n^3-12n^2+47n-60\right)\)
\(=\left(n-2\right)\left(n^3-3n^2-9n^2+27n+20n-60\right)\)
\(=\left(n-2\right)\left[n^2\left(n-3\right)-9n\left(n-3\right)+20\left(n-3\right)\right]\)
\(=\left(n-2\right)\left(n-3\right)\left(n^2-9n+20\right)\)
\(=\left(n-2\right)\left(n-3\right)\left(n-4\right)\left(n-5\right)\)
Vì n-2;n-3;n-4 và n-5 là bốn số nguyên liên tiếp nên
\(\left(n-2\right)\left(n-3\right)\left(n-4\right)\left(n-5\right)⋮24\)
hay \(A⋮24\)(đpcm)
Câu ||
ĐKXĐ x≥-2
\(\Rightarrow x^2+2x+2=\left(2x+1\right)\sqrt{x+2}\)
\(\Leftrightarrow x^2+2x+2-2x\sqrt{x+2}-\sqrt{x+2}=0\)
\(\Leftrightarrow x^2-2x\sqrt{x+2}+x+2+\left(x-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+2}\right)^2+\left(x-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+2}\right)\left(x-\sqrt{x+2}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+2}=0\left(1\right)\\x-\sqrt{x+2}+1=0\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow x=\sqrt{x+2}\) \(\Leftrightarrow x^2=x+2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Từ (2) \(\Rightarrow\sqrt{x+2}=-\left(x+1\right)\) (do vế trái ≥ 0 nên -(x+1)≥0 ⇒x≤-1 kết hợp điều kiện là x≥-2 ⇒\(-2\le x\le-1\) )\(\Leftrightarrow x^2+x-1=0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{2}\left(L\right)\\x=\dfrac{-\sqrt{5}-1}{2}\left(TM\right)\end{matrix}\right.\)
Vậy...
Câu |||
Vì VP=2022⋮2, \(20x^2⋮2\) \(\Rightarrow13y^2⋮2\) Mà (13;2)=1 \(\Rightarrow y^2⋮2\Rightarrow y⋮2\Rightarrow y=2t\left(t\in Z\right)\)
pt trở thành \(20x^2+13\left(2t\right)^2=2022\Rightarrow10x^2+26t^2=1011\)
\(\Rightarrow2\left(5x^2+13t^2\right)=1011\) Vô lí vì VT⋮2 mà VP\(⋮̸\)2 ⇒ ko tồn tại x;t ∈Z
⇒ko tồn tại x;y∈Z Vậy ...
