\(TH1:x\le1\)

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

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

\(x=\frac{5}{4}\left(KTM\right)\)

\(TH2:1< x\le3\)

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

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

\(x=\frac{3}{2}\left(TM\right)\)

\(TH3:x>3\)

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

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

\(0x=3\left(KTM\right)\)

vậy nghiệm duy nhất của pt là \(x=\frac{3}{2}\)

Ta có (ac + bd)² - (ad + bc)² 

 =(ac + bd + ad + bc))(ac + bd - ad - bc) 

= [c(a + b) + d(a + b)][c(a - b) - d(a - b)]

= (c + d)(a + b)(a - b)(c - d) 

 =(a² - b²)(c² - d²)  (đpcm)

Ta có a³ + b³ + c³ - 3abc = (a + b)³ - 3ab(a + b) + c³ - 3abc 

= [(a + b)³ + c³] - [3ab(a + b) + 3abc]

= (a + b + c)[(a + b)² - (a + b)c + c²] - 3ab(a + b + c) 

= (a + b + c)(a² + b² + c²  + 2ab - ac - bc - 3ab) 

= (a + b + c)(a² + b² + c²  - ac - bc - ab)  

Khi đó 2(a³ + b³ + c³ - 3abc) = 2(a + b + c)(a² + b² + c²  - ac - bc - ab)  

= (a + b + c)(2a² + 2b² + 2c² - 2ac - 2bc - 2ab) 

= (a + b + c)[(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²)] 

= (a + b + c)[(a - b)² + (b - c)² + (c - a)²] (đpcm)

ta có :

\(S=-x^2+2xy-4y^2+2x+10y-8=-\left(x-y\right)^2-3y^2+2x+10y-8\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-3y^2+12y-8=-\left(x-y-1\right)^2-3\left(y^2-4y+4\right)+5\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\)

Dấu = xảy ra khi \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

\(\frac{\left(135^{^2}\right)+130.135+65^2}{135^2-65^2}\)

=\(\frac{135^2+2.65.135+65^2}{\left(135-65\right).\left(135+65\right)}\)

=\(\frac{\left(135+65\right)^2}{70.200}\)

=\(\frac{200.200}{70.200}\)

=\(\frac{20}{7}\)

Học tốt

T cho mk nhé

Dài thế:))