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1/
a,\(A=x-x^2=-x^2+x=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow A=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra <=>x=1/2
Vậy Amax=1/4 khi x=1/2
b, \(B=2x-2x^2-5=-2x^2+2x-5\)
\(\Rightarrow2B=-4x^2+4x-10=-\left(4x^2-4x+1\right)-9=-\left(2x-1\right)^2-9\)
Vì \(-\left(2x-1\right)^2\le0\Rightarrow2B=-\left(2x-1\right)^2-9\le-9\Rightarrow B\le\frac{-9}{2}\)
Dấu "=" xảy ra <=>x=1/2
Vậy Bmax=-9/2 khi x=1/2
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\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
a) (2x+1)^2+2(4x^2-2)+(2x-1)^2=4x2+4x+1+8x2-4+4x2-4x+1=16x2-2
4( 3x + 4 )2 + 2x( 2x - 5 )( 2x + 5 ) - ( 2x + 3 )2
= 4( 9x2 + 24x + 16 ) + 2x( 4x2 - 25 ) - ( 4x2 + 12x + 9 )
= 36x2 + 96x + 64 + 8x3 - 50x - 4x2 - 12x - 9
= 8x3 + ( 36x2 - 4x2 ) + ( 96x - 50x - 12x ) + ( 64 - 9 )
= 8x3 + 32x2 + 34x + 55
Ta luôn có \(a^2+b^2-2ab=\left(a-b\right)^2\)
Áp dụng hằng đẳng thức trên thì có :
\(\left(2x+3\right)^2+\left(2x+5\right)^2-2\left(2x+3\right)\left(2x+5\right)\)
\(=\left[\left(2x+5\right)-\left(2x+3\right)\right]^2\)
\(=2^2\)
\(=4\)
\(\left(2x+3\right)^2+\left(2x-5\right)^2-2\left(2x+3\right)\left(2x-5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x-5\right)\right]^2\)
\(=\left(2x+3-2x+5\right)^2=8^2=64\).
Merry Chrismas.
`@` `\text {Ans}`
`\downarrow`
`(2x - 3)^2 - (2x + 3)^2`
`= 4x^2 - 12x + 9 - (4x^2 + 12x + 9)`
`= 4x^2 - 12x + 9 - 4x^2 - 12x - 9`
`= (4x^2 - 4x^2) + (-12x - 12x) + (9-9)`
`= -24x`
____
`@` CT:
`(A + B)^2 = A^2 + 2AB + B^2`
`(A - B)^2 = A^2 - 2AB + B^2`
\(\left(2x-3\right)^2-\left(2x+3\right)^2\)
\(=\left[\left(2x-3\right)+\left(2x+3\right)\right]\left[\left(2x-3\right)-\left(2x+3\right)\right]\)
\(=\left(2x-3+2x+3\right)\left(2x-3-2x-3\right)\)
\(=4x\cdot-6\)
\(=-24x\)