\(25a^2-25x^2+10x-1\)

\(=25a^2-\left(5x-1\right)^2\)

\(=\left(5a-5x+1\right)\left(5a+5x-1\right)\)

\(25a^2-25x^2+10x-1=25a^2-\left(25x^2-10x+1\right)=\left(5a\right)^2-\left(5x-1\right)^2=\left(5a+5x-1\right)\left(5a-5x+1\right)\)

\(x^2-25=\left(x-5\right)\left(x+5\right)\\ x^2+10x+25=\left(x+5\right)^2\\ x^2-6x+xy-6y=x\left(x-6\right)+y\left(x-6\right)=\left(x+y\right)\left(x-6\right)\\ x^2-2x-y^2+1=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

1. 10x²(x - 1) + 15xy(1 - x)

= 10x²(x - 1) - 15xy(x - 1)

= (10x² - 15xy)(x - 1)

= 5x(2x - 3y)(x - 1)

2. Hình như sai đề

2. 25x² - x² + 10x - 2

= 25x² + 10x - x² - 2

= -(25x² - 10x + x²) - 2

= -(5x - x)² - 2

= -(5x - x)² - \(\left(\sqrt{2}\right)^2\)

= -(5x + x + \(\sqrt{2}\))(5x + x - \(\sqrt{2}\))

\(25x^2-10x+1-16z^2=\left(5x-1-4z\right)\left(5x-1+4z\right)\)

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

\(F=-x^2+2x-1+1\)

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

Dấu = xảy ra khi x=1

\(A=1-25x^2-10x\)

\(=-\left(25x^2+10x-1\right)\)

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

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

A = 1 - 25x^2 - 10x = - ( 25x^2 + 10x - 1 )

= - [ ( 5x + 1 ) ^ 2 - 2 ]

= 2 - ( 5x + 1 ) ^2

`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

`<=> -5x+5=0`

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

`3,(3x+1)^2-(2x+1/2)^2=0`

`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`

`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

`<=>(x+4)^2=0`

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

`<=> (5-x)^2=0`

`<=>5-x=0`

`<=>x=5`

a/ 2x - 1/4 = 0

<=> 2x      = 1/4

<=> x        = 1/8

Vậy: S = {1/8}

b/ 25x^2 - 10x = 0

<=> 5x ( 5x - 2 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}5x=0\\5x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\5x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\frac{2}{5}\end{cases}}}\)

Vậy: S = {0;2/5}

`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`