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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Rút gọn các biểu thức:
i) (x + 5)(x – 5) – (x2 – 1)
= x2 - 25 - x2 + 1
= -24
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=5x-10+2x+6=7x-4\\ b,=x^2+2x+1-x^2+3x+10=5x+11\\ c,=x^2-49-x^2+1=-48\\ d,\text{Đề có sai ko vậy?}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đkxđ:\(x\ne5,x\ne-5\)
\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5x+25}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)
thay x=1 vào bt A, ta được:
\(-\dfrac{4}{1-5}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với `x \ne -5,x \ne -1` có:
`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`
`A=[x^2-3x-4]/[(x+5)(x+1)]`
`A=[(x-4)(x+1)]/[(x+5)(x+1)]`
`A=[x-4]/[x+5]`
\(=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+x+5x+5}-\dfrac{1}{x+1}\\ =\dfrac{x+2}{x+5}+\dfrac{-5x-1}{\left(x^2+x\right)+\left(5x+5\right)}-\dfrac{1}{x+1}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{x\left(x+1\right)+5\left(x+1\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{\left(x+1\right)\left(x+5\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+2x+x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+x-4x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x^2+x\right)-\left(4x+4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x\left(x+1\right)-4\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x-4}{x+5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`
ii) (5x + 1)2 + (5x – 1)2 + 2(5x + 1)(5x – 1)
= [(5x + 1) + (5x - 1)]2
= (10x)2 = 100x2