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`-10x(2-x)-5x(x+2)=5x(x+3)`
`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`
`<=> 5x^2 - 30x = 5x^2 + 15x`
`<=> -30x = 15x`
`<=> -45x = 0`
`<=> x = 0`
Vậy `S = {0}`
\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)
\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)
\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)
\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)
\(\text{⇔}15x^2-30=-5x^2+15\)
\(\text{⇔}15x^2+5x^2=30+15\)
\(\text{⇔}20x^2=45\)
\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
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\(5x^4y+10x^3y+10x^2y^3+5xy^4\)
\(=5xy.x^3+5xy.2x^3+5xy.2xy^3+5xy.y^3\)
\(=5xy\left(x^3+2x^3+2xy^3+y^3\right)\)
Ht pt
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\(=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}\cdot\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)
\(\dfrac{x^2+x}{5x^2+10x+5}:\dfrac{3x+3}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x^2+2x+1\right)}:\dfrac{3\left(x+1\right)}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}.\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)
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\(a.\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8.\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
\(\Leftrightarrow-4x+3=8\)
\(\Leftrightarrow-4x=5\)
\(\Leftrightarrow x=-\frac{5}{4}\)
\(b.\left(3x-5\right)\left(7-5x\right)+\left(5x-2\right)\left(3x-2\right)-2=0\)
\(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
\(\Leftrightarrow x=\frac{41}{42}\)
=5x.x^2-5x.2x+5x.1 =5x(x^2-2x+1)