\(x^2-6x+8=0\\ \Leftrightarrow x^2-4x-2x+8=0\\ \Leftrightarrow x\left(x-4\right)-2\left(x-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy đa thức \(x^2-6x+8\) có 2 nghiệm số \(2\) và \(4\)

\(f\left(0\right)=5\\ \Leftrightarrow a\cdot0^2+b\cdot0+c=c=5\\\Rightarrow c=5\\ f\left(1\right)=3\\ \Leftrightarrow a\cdot1^2+b\cdot1+c=a+b+5=3\\ \Leftrightarrow a+b=-2\\ \Leftrightarrow2a+2b=-4\\ f\left(-2\right)=4\\ \Leftrightarrow a\cdot\left(-2\right)^2+b\cdot\left(-2\right)+c=4a-2b+5=4\\ \Leftrightarrow4a-2b=-1\\ 2a+2b+4a-2b=-4+\left(-1\right)\\ \Leftrightarrow6a=-5\\ \Leftrightarrow a=\dfrac{-5}{6}\\ a+b=-2\\ \Leftrightarrow\dfrac{-5}{6}+b=-2\\ \Leftrightarrow b=\dfrac{-7}{6}\)

900

li ke nha Minamoto Shuki

Đặt \(z = a + bi (a,b \in \mathbb{Z})\)

Ta có:

\(z^2+\left|z\right|=0\\ \Leftrightarrow\left(a+bi\right)^2+\left|a+bi\right|=0\\ \Leftrightarrow a^2-b^2+2abi+\sqrt{a^2+b^2}=0+0i\\ \Leftrightarrow\left\{{}\begin{matrix}2ab=0\left(1\right)\\a^2-b^2+\sqrt{a^2+b^2}=0\left(2\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\\\text{Nếu }a=0\\ \Rightarrow\left(2\right)\Leftrightarrow\left|b\right|-b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=1\\b=-1\end{matrix}\right.\\ \text{Nếu }b=0\\ \Rightarrow\left(2\right)\Leftrightarrow\left|a\right|+a^2=0\\ \Leftrightarrow a=0\)

Vậy

\(\left(a,b\right)\in\left\{\left(0;0\right);\left(0;1\right);\left(0;-1\right)\right\}\\ \Rightarrow z\in\left\{0;i;-i\right\}\)

\(81^7 - 27^9 - 9^{13}\\ = (3^4)^7 - (3^3)^9 - (3^2)^{13} \\ = 3^{4.7} - 3^{3.9} - 3^{2.13} \\ = 3^{28} - 3^{27} - 3^{26} \\ = 3^{24}(3^4-3^3-3^2) \\ = 3^{24}(81-27-9) \\ =3^{24} . 45 \vdots 45 \)

\(10^9+10^8+10^7\\=10^6(10^3+10^2+10)\\=10^6(1000+100+10)\\=10^6 . 1110 \\ =10^6 . 5 .222\vdots 222\)

\(29\cdot\left(19-13\right)-19\cdot\left(29-13\right)\\ =29\cdot19-29\cdot13-19\cdot29+19\cdot13\\ =\left(29\cdot19-19\cdot29\right)+\left(19\cdot13-29\cdot13\right)\\ =0+13\left(19-29\right)\\ =13\cdot\left(-10\right)\\ =-130\)

\(a\left(a+2\right)+b\left(b-2\right)-2ab=63\\ \Leftrightarrow a^2+2a+b^2-2b-2ab=63\\ \Leftrightarrow a^2-2ab+b^2+2a-2b+1=64\\ \Leftrightarrow\left(a-b\right)^2+2\left(a-b\right)+1=64\\ \Leftrightarrow\left(a-b+1\right)^2=64\\ \Leftrightarrow\left[{}\begin{matrix}a-b+1=8\\a-b+1=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}a-b=9\\a-b=-7\end{matrix}\right.\)

Vậy ...