#)Giải :

\(9^8:\left(27^3.81^2\right)=\left(3^2\right)^8:\left[\left(3^3\right)^3.\left(3^4\right)^2\right]=3^{16}:\left[3^9.3^8\right]\)

Đến đây ez rùi ^^

9⁸:(27³.81²) = 9⁸:(3^3x3.3^4x2)=9⁸:3¹⁷= \(\frac{1}{3^9}\)

\(9^8:\left(27^3.81^2\right)=3^{16}:\left(3^9.3^8\right)=3^{16}:\left(3^{9+8}\right)=3^{16}:3^{17}=3^{-1}=\frac{1}{3}\)

\(9^8:\left(27^3.81^2\right)\)

\(=3^{16}:\left(3^9.3^8\right)\)

\(=3^{16}:3^{17}\)

\(=\frac{1}{3}\)

\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

\(\Leftrightarrow\) ( x - 1 ).( x + 3 ) = ( x + 2 ).( x - 2 )

\(\Leftrightarrow\) x² + 2x - 3 = x² - 4

\(\Leftrightarrow\)x² - x² + 2x = -4 + 3

\(\Leftrightarrow\) 2x = -1

\(\Leftrightarrow\) x = \(\frac{-1}{2}\)

Vậy \(x=\frac{-1}{2}\)

\(\Leftrightarrow\frac{1}{1+x^2}-\frac{1}{1+xy}+\frac{1}{1+y^2}-\frac{1}{1+xy}\ge0.\)

\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\frac{x\left(y-x\right)\left(1+y^2\right)+y\left(x-y\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(y+x^2y-x-xy^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\frac{\left(x-y\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\left(lđ\forall x,y\ge1\right)\)

Dấu "=" xra khi x=y=1

có 19 số tự tự nhiên giữa chúng nên hiệu là :

                19 + 1 = 20 

số lớn là : ( 520 + 20 ) : 2 = 270 

Vậy số lớn là 270

\(\left(\sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}\right)^2\le\left(1^2+1^2+1^2\right)\left(4x+1+4y+1+4z+1\right)=21.\)

\(\Leftrightarrow\sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}\le\sqrt{21}\left(đpcm\right)\)

Dấu "=" xra :

\(\frac{4x+1}{1}=\frac{4y+1}{1}=\frac{4z+1}{1}\Rightarrow x=y=z=\frac{1}{3}\)

4x⁴ + 20x² + 25

= ( 2x²)² + 2.2.x².5 + 5²

= ( 2x² + 5 )²

TL:

\(4x^4+20x^2+25\)  

=\(\left(2x^2\right)^2+2.2.x^2.5+25\)

\(=\left(2x^2+5\right)^2\) 

hc tốt

#)Giải :

Điệu kiện : \(x\in Z;x< 0\)

Thử đi bạn, số nào cg ra 

\(x^3< x^2\)

\(\Rightarrow x.x.x< x.x\)

tức x thuộc số âm

\(\Rightarrow x\in z,x< 0\)

XXX   x   X  = X

000 x 0 = 0 

Vậy X = 0

xxx*x=...x

=> x=1 hoặc x=0

ko cần k chúc bạn học tốt