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a) = x^2 + 2x + 1 - x^2 +2x - 1 -3x^2 +x - x - 1
= - 3x^2 +4x -1
b) =5x^2 + 10x - 10x - 20 - 1/2 .(36 - 96x + 64x^2 ) +17
= 5x^2 - 20 - 18 - 48 x - 32x^2 +17
= -27x^2 - 48x - 3
Chúc bn hok tốt a !
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
a) = \(12a^2b\left(a^2-b^2\right)\)
= \(12a^4b-12a^2b^3\)
b)nhân ra :
= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)
= \(2x^4-19x^3+33x^2-46x+10\)
Tìm x:
a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)
= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)
=\(2x=-14=>x=-7\)
b) \(x^3+27-x\left(x^2-1\right)=27\)
= \(x^3+27-x^3+x=27\)
= \(27+x=27=>x=0\)
\(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\frac{1}{2}\left(36-96x+64x^2\right)+17\)
\(=5x^2-20+30-32x^2+17=-27x^2+27\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
a) (x+1)2(x-1)2 - 3(x+1)(x-1)
= (x+1)(x-1)[(x+1)(x-1) - 3 ]
= (x+1)(x-1)[ x2 -4]
=(x+1)(x-1)(x-2)(x+2)