\(25x^2-9=0\\ \Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=3\\5x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

=>(5x-3)(5x+3)=0

=>5x-3=0 hoặc 5x+3=0

=>x=3/5 hoặc x=-3/5

`-25x^{2}+(y^{2}-6y+9)`

`=-25x^{2}+(y-3)^{2}`

`=(y-3)^{2}-(5x)^{2}`

`=(y-3-5x)(y-3+5x)`

Ta có: \(\left(y^2-6y+9\right)-25x^2\)

\(=\left(y-3\right)^2-25x^2\)

\(=\left(y-3-5x\right)\left(y-3+5x\right)\)

a) Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)=35\)

\(\Leftrightarrow x^3+8=35\)

\(\Leftrightarrow x^3=27\)

hay x=3

b) Ta có: \(\left(25x^2+5x+1\right)\left(5x-1\right)=-9\)

\(\Leftrightarrow125x^3-1=-9\)

\(\Leftrightarrow125x^3=-8\)

\(\Leftrightarrow x=-\dfrac{2}{5}\)

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

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

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

x2 nghĩa là x^2 hả bạn?

Giúp mik vs

\(a,=3\left(x-5\right)-x\left(x-5\right)=\left(3-x\right)\left(x-5\right)\\ b,=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\\ c,=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\\ d,=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-25x^2=\left(y-5x-3\right)\left(y+5x-3\right)\)

Với x=2 thì \(25\cdot2-30x+9\)trở thành:

\(50-60+9=-1\)

25x²-30x+9

Thay x=2 vào biểu thức trên, ta có :

   25.2²-30.2+9=49

Vậy:.....

#H

25x2 - 30x + 9

= 2522 - 302 + 9 

= 2220 + 9 

= 2229 

a) Ta có 3 x − 1 10 x 2 + 2 x . 25 x 2 + 10 x + 1 1 − 9 x 2 = − 5 x + 1 2 x ( 3 x + 1 )  

b) Kết quả  = p . ( p − 3 ) 7