\(3x^2+14x-15=0\)

\(\Delta=b^2-4ac=14^2-4\cdot3\cdot15=16\)

=> Phương trình có 2 nghiệm

\(=>\orbr{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-14+\sqrt{16}}{2\cdot3}=\frac{-10}{6}=\frac{-5}{3}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-14-\sqrt{16}}{2\cdot3}=\frac{-18}{6}=-3\end{cases}}\)

Vậy...

Sai r babe :))

\(3x^2+14x-15=0\)

\(\Delta=14^2-4.3.\left(-15\right)=376\)

\(x_1=\frac{-14-\sqrt{376}}{28};x_2=\frac{-14+\sqrt{375}}{28}\)

a. \(x\left(x+2\right)\left(x+1\right)\left(x-1\right)\)

3x - 3y + x² - y²

= 3(x - y) + (x - y)(x + y)

= (x - y)(x + y + 3)

\(3x-3y+x^2-y^2\)

\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y+3\right)\)

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

\(\Leftrightarrow x^2-x-x^2+2x=5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

Ta có: \(a^2+b=b^2+c\Rightarrow a^2-b^2=c-b\Rightarrow\left(a+b\right)\left(a-b\right)-\left(a-b\right)=c-a\Rightarrow\left(a+b-1\right)\left(a-b\right)=c-a\)(1)\(b^2+c=c^2+a\Rightarrow b^2-c^2=a-c\Rightarrow\left(b+c\right)\left(b-c\right)-\left(b-c\right)=a-b\Rightarrow\left(b+c-1\right)\left(b-c\right)=a-b\)(2)\(c^2+a=a^2+b\Rightarrow c^2-a^2=b-a\Rightarrow\left(c+a\right)\left(c-a\right)-\left(c-a\right)=b-c\Rightarrow\left(c+a-1\right)\left(c-a\right)=b-c\)(3)

Nhân ba vế của ba đẳng thức (1), (2), (3), ta được:\(\left(a+b-1\right)\left(a-b\right)\left(b+c-1\right)\left(b-c\right)\left(c+a-1\right)\left(c-a\right)=\left(c-a\right)\left(a-b\right)\left(b-c\right)\Rightarrow\left(a+b-1\right)\left(b+c-1\right)\left(c+a-1\right)=1\)(Do a, b, c đôi mội khác nhau nên \(a-b,b-c,c-a\ne0\) )

Ta có : 

a² + b = b² + c <=> a² - b² = c - b <=> ( a + b ) ( a - b ) = c - b

<=> \(a+b=\frac{c-b}{a-b}\)<=> \(a+b-1=\frac{c-a}{a-b}\)

b² + c = c² + a <=> b² - c² = a - c <=> ( b + c ) ( b - c ) = a - c

<=> \(b+c=\frac{a-c}{b-c}\)<=> \(b+c-1=\frac{a-b}{b-c}\)

a² + b = c² + a <=> a² - c² = a - b <=> ( a + c ) ( a - c ) = a - b

<=> \(a+c=\frac{a-b}{a-c}\)<=> \(a+c-1=\frac{c-b}{a-c}\)

Suy ra :

( a + b - 1 ) ( a + c - 1 ) ( a + c - 1 ) = \(\frac{c-a}{a-b}.\frac{a-b}{b-c}.\frac{c-b}{a-c}=-\frac{a-c}{a-b}.\frac{a-b}{b-c}.\left(-\frac{b-c}{a-c}\right)=1\)