t mink

1+1=2

2+2=4

3+3=6

123456789+987654321=1111111110

\(A=2x^2+y^2-2xy+x+2\)

\(A=\left(x^2-2xy+y^2\right)+\left[x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{7}{4}\)

\(A=\left(x-y\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\\left(x+\frac{1}{2}\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x-y\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{7}{4}=A\ge\frac{7}{4}>0\forall x;y\)

Vậy không có các số tự nhiên thỏa mã đẳng thức \(A=2x^2+y^2-2xy+x+2=0\)

2⁴ . 6¹⁵ . 3⁴

= 2⁴ . ( 2.3)¹⁵ . 3⁴

=2⁴ . 2¹⁵ . 3 ¹⁵ . 3⁴

= 2¹⁹ . 3¹⁹

= 6¹⁹

\(5^{13}.30^{21}.6^{17}=5^{13}.5^{21}.6^{21}.6^{17}\)\(=5^{34}.6^{38}\)\(=5^{34}.6^{34}.6^4\)\(=30^{34}.6^4\)

huhu , ai rảnh thì làm giúp mk vs ạ !!!

a)  \(x^3+8x^2+17x+10\)

\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+7x+10\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

b)  \(=x^4-2x^3-12x^2+12x+36\)

\(=x^2\left(x^2-2x-6\right)-2\left(x^2-2x-6\right)\)

\(=\left(x^2-2\right)\left(x^2-2x-6\right)\)

Giúp mk vs ạ , mk đang cầng gấp !!!

x^m+3+1+x^m+3-(x+1)=x^m+3x+x^m+3-(x+1)=x^m+3(x+1)-(x+1)=(x+1)(x^m+3-1)

 Ta có : x^m+4 + x^m+3,-x-1

<=>x^m. x⁴ + x^m . x³ - (x+1)

<=> x^m+3. (x+1) -( x+1)

<=> (x^m+3-1)(x+1)