Đáp án+Giải thích các bước giải:
Bài 1:
`a)6x^2-4x^2(x^2-3x-2)`
`=6x^2-4x^4+12x^3+8x^2`
`=-4x^4+12x^3+14x^2`
`b)(3x-2)^2-2(x-1)^2`
`=9x^2-12x+4-2x^2+4x-2`
`=7x^2-8x+2`
`c)(2x^4+5x+x^3-2-3x^2):(x^2+1-x)`
`=[(2x^4+2x^2-2x^3)+(3x^3+3x-3x^2)-(2x^2+2-2x]:(x^2+1-x)`
`=[2x^2(x^2+1-x)+3x(x^2+1-x)-2(x^2+1-x)]:(x^2+1-x)`
`=(x^2+1-x)(2x^2+3x-2):(x^2+1-x)`
`=2x^2+3x-2`
Bài 2:
`a)4a^2-9=(2a-3)(2a+3)`
`b)3/5 x^3 y^2-9/4 x^2 y^3-6xy`
`=6xy(1/10x^2y-3/8xy^2-1)`
`c)x^3-2x^2-9x+18`
`=(x^3-2x^2)-(9x-18)`
`=x^2(x-2)-9(x-2)`
`=(x-2)(x^2-9)`
`=(x-2)(x-3)(x+3)`
`d)2x^2-3x+1`
`=2x^2-2x-x+1`
`=(2x^2-2x)-(x-1)`
`=2x(x-1)-(x-1)`
`=(x-1)(2x-1)`
Bài 3:
`a)3x^3-75x=0`
⇔`3x(x^2-25)=0`
⇔`3x(x-5)(x+5)=0`
⇔\(\left[ \begin{array}{l}3x=0\\x-5=0\\x+5=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=5\\x=-5\end{array} \right.\)
Vậy `S={0;±5}`
`b)(3x-4)^2-9(x-1)(x-3)=13`
⇔`9x^2-24x+16-9(x^2-4x+3)=13`
⇔`9x^2-24x+16-9x^2+36x-27-13=0`
⇔`12x-24=0`
⇔`12x=24`
⇔`x=2`
Vậy `S={2}`
Bài 4:
Ta có:`x^2-x+1/3`
`=(x^2-2.x. 1/2+1/4)+1/12`
`=(x-1/2)^2+1/12>0`
Vì `(x-1/2)^2≥0` với mọi `x`
`1/12>0`
⇒`x^2-x+1/3>0` với mọi số thực `x`(Đpcm)
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Chúc bạn học tốt!!!.
`#Rùa ~ ~ ~`
