25a2-25x2+10x-1
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\(25a^2-25x^2+10x-1\)
\(=25a^2-\left(5x-1\right)^2\)
\(=\left(5a-5x+1\right)\left(5a+5x-1\right)\)
\(x^2-25=\left(x-5\right)\left(x+5\right)\\ x^2+10x+25=\left(x+5\right)^2\\ x^2-6x+xy-6y=x\left(x-6\right)+y\left(x-6\right)=\left(x+y\right)\left(x-6\right)\\ x^2-2x-y^2+1=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
\(25x^2-10x+1-16z^2=\left(5x-1-4z\right)\left(5x-1+4z\right)\)
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
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}
\(=25a^2-\left(5x-1\right)^2=\left(5a+5x-1\right)\left(5a-5x+1\right)\)
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