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a, 25^2 - 15^2 = ( 25 - 15 )( 25 + 15) = 10 . 40 = 400
b, 87^2 + 73^2 - 27^2 - 13^2
= 87^2 - 27^2 + 73^2 - 13^2
= ( 87 - 27)( 87 + 27) + (73 - 13 )(73+ 13)
= 60 . 114 + 60 . 86
= 60( 114 + 86)
= 60 .200
= 12000
c, x^3 + 27 + 9 x^2 + 27x
= x^3 + 27x + 9x^2 + 27
=(x + 3)^3
thay x =97 ta có
= (97 + 3)^3
= 100^3
=1000000
d, 1,6^2 + 4.0,8.3,4 + 3,4^2 ( nè 3,4^2 chứ không phải 3,42)
= 1,6^2 + 2.2.0,8.3,4 + 3,4^2
=1,6^2 + 2.1,6.3,4 + 3,4^2
= (1,6 + 3,4)^2
= 5^2
= 25
e, x = 11 => 12 =x + 1 thay vào ta có
x^4 - ( x+ 1)x^3 + (x+1)x^2 -(x+1)x + 11
= x^4 - x^4 - x^13 + x^3 + x^2 - x^2 - x + 11
= -x + 11
= -11 + 11
= 0
ĐÚng ch o tui nha
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(x^2+6x+9=\left(x+3\right)^2\)
d) \(4x^2+4x+1=\left(2x+1\right)^2\)
e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)
f) \(4x^2+12x+9=\left(2x+3\right)^2\)
g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)
h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)
a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)2
b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)2
c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)2
d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)2
\(a,A=-x^2-6x-10=-\left(x^2+6x+9\right)-1=-\left(x+3\right)^2-1\le-1\)
Dấu = xảy ra ⇔ x +3 =0 ⇔ x = -3
\(Max_A=-1\text{ ⇔}x=-3\)
\(b,B=12x-4x^2+3=-\left(4x^2-12x+9\right)+12=-\left(2x-3\right)^2+12\le12\)
Dấu = xảy ra \(\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
\(Max_B=12\text{ ⇔}x=\dfrac{3}{2}\)
\(c,8x-8x^2+3=-8\left(x^2-x+\dfrac{1}{4}\right)+5=-8\left(x-\dfrac{1}{2}\right)^2+5\le5\)
\(d,-x^2-8x+2018-y^2+4y\)
\(=-\left(x^2+8x+16\right)-\left(y^2-4y+4\right)+2038\le2038\)
\(e,-4x^4-12x^2+11=-\left(4x^4+12x^2+9\right)+20=-\left(2x^2+3\right)^2+20\le20\)
\(f,C=x-\dfrac{x^2}{4}\Rightarrow4C=4x-x^2\)\(=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\)
\(\Rightarrow C=-\dfrac{\left(x-2\right)^2}{4}+1\le1\)
\(g,D=x-\dfrac{9x^2}{25}\Rightarrow25D=-\left(9x^2-25x\right)=-\left(9x^2-2.3x.\dfrac{25}{6}+\dfrac{625}{36}\right)+\dfrac{625}{36}=-\left(3x-\dfrac{25}{6}\right)^2+\dfrac{625}{36}\)
\(\Rightarrow D=\dfrac{-\left(3x-\dfrac{25}{6}\right)^2}{25}+\dfrac{25}{36}\le\dfrac{25}{36}\)
\(4x^2-4x-5=4x^2-4x+1-6=\left(2x-1\right)^2-6\ge-6\)
\(Min=-6\Leftrightarrow x=\dfrac{1}{2}\)
\(4x^2+12x+10=4\left(x^2+3x+\dfrac{9}{4}\right)+1=4\left(x+\dfrac{3}{2}\right)^2+1\ge1\)
\(Min=1\Leftrightarrow x=-\dfrac{3}{2}\)
\(4x^2-12x-5=4\left(x^2-3x+\dfrac{9}{4}\right)-14=4\left(x-\dfrac{3}{2}\right)^2-14\ge-14\)
\(Min=-14\Leftrightarrow x=\dfrac{3}{2}\)
\(9x^2+12x+8=\left(9x^2+12x+4\right)+4=\left(3x+2\right)^2+4\ge4\)
\(Min=4\Leftrightarrow x=-\dfrac{2}{3}\)
do hơi bận nên mk ghi đáp án nha, ko hiểu đâu ib mk
a) \(3xy^2-2xy+12x=x\left(3y^2-2y+12\right)\)
b) \(x^3-10x^2+25x-16xy^2=x\left(x-4y-5\right)\left(x+4y-5\right)\)
c) \(5y^3-10xy^2+5x^2y-20y=5y\left(y-x-2\right)\left(y-x+2\right)\)
d) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)\left(x+y-z\right)\)
e) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
f) \(8-12x+6x^2-x^3=\left(2-x\right)^3\)
g) \(125x^3-75x^2+15x-1=\left(5x-1\right)^3\)
h) \(x^2-xz-9y^2+3yz=\left(x-3y\right)\left(x+3y-z\right)\)
