\(-6x^2+x+1=0\)

\(\Leftrightarrow-6x^2+3x-2x+1=0\)

\(\Leftrightarrow-\left(6x^2-3x\right)-\left(2x-1\right)=0\)

\(\Leftrightarrow-3x\left(2x-1\right)-\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(-3x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{3}\end{cases}}\)

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

-6x2 + x+1 = 0

  -12 + x+1 = 0

  - 12+ x     = 0- 1

 - 12 +x     = -1 

          x     = -1  -  -12

          x     = -11

Vậy x = -11

dễ :>>>

\(-14x=10\)

\(x=10:\left(-14\right)\)

\(x=\frac{-5}{7}\)

P=(5-x)(x+5)+(x-3)^2+5x

=25-x^2+x^2-6x+9+5x

=-x+34

a, \(\left(x-8\right)^2-9^2=\left(x-8-9\right)\left(x-8+9\right)=\left(x-17\right)\left(x+1\right)\)

b, \(16-x^2+4xy-4y^2=4^2-\left(x^2-4xy+4y^2\right)=4^2-\left(x-2y\right)^2\)

\(=\left(4-x+2y\right)\left(4+x-2y\right)\)

a 

( x - 8 ) ^ 2 - 81 

= ( x - 8 ) ^ 2 - 9 ^ 2 

= ( x - 8 - 9 ) ( x - 8 + 9 ) 

= ( x - 17 ) ( x + 1 ) 

b 

16 - x^2 + 4xy - 4y^2 

= 4^2 - ( x^2 - 4xy + 4y^2 ) 

= 4^2 - ( x - 2y ) ^ 2 

= [ 4 - ( x - 2y ) ] [ 4 + ( x - 2y ) ] 

= ( 4 - x + 2y ) ( 4 + x - 2y ) 

Ta có x + y + z = 0

=> x + y = -z

y + z = -x

x + z = -y

Khi đó Q = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}=\frac{-z}{y}.\frac{-x}{y}.\frac{.-y}{x}=-1\)

Vậy khi x + y + z = 0 thì Q = -1

ta có 

\(Q=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)

Áp dụng bất đẳng thức Bunhiacopxki, ta có:
VT.[a(b+c)+b(c+d)+c(d+a)+d(a+b)]≥(a+b+c+d)²
Ta cần chứng minh:
(a+b+c+d)²≥2(ab+bc+cd+da+2ca+2bd)

⇔a²+b²+c²+d²≥2ca+2bd⇔(a−c)²+(b−d)²≥0
Bất đẳng thức đã được chứng minh

dấu = xảy ra khia a=c, b=d

Bc nhé = 35