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pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
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a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
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\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)
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`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
Em lớp 5 nên giải đc câu đầu thôi ạ :))
\(\frac{x-25}{1990}+\frac{x-29}{1986}+\frac{x+1}{1008}=4\)
\(\Leftrightarrow\left(\frac{x-25}{1990}-1\right)+\left(\frac{x-29}{1986}-1\right)+\left(\frac{x+1}{1008}-2\right)=0\)
\(\Leftrightarrow\frac{x-2015}{1990}+\frac{x-2015}{1986}+\frac{x-2015}{1008}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\frac{1}{1990}+\frac{1}{1986}+\frac{1}{1008}\right)=0\)
\(\Leftrightarrow x-2015=0\)
\(\Leftrightarrow x=2015\)
\(\left(2x-5\right)^3-\left(x-2\right)^3=\left(x-3\right)^3\)
\(\Leftrightarrow\left(2x-5-x+2\right)^3+3\left(2x-5\right)\left(x-2\right)\left(2x-5-x+2\right)=\left(x-3\right)^3\)
\(\Leftrightarrow\left(x-3\right)^2+3\left(2x-5\right)\left(x-2\right)\left(x-3\right)=\left(x-3\right)^3\)
\(\Leftrightarrow\left(x-3\right)^2+3\left(2x-5\right)\left(x-2\right)\left(x-3\right)-\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)\left(\left(x-3\right)+3\left(2x-5\right)\left(x-2\right)-\left(x-3\right)^2\right)=0\)
Đến đây bắt buộc nhân đa thức với đa thức rồi tính thôi :v , tự tính nha