Giải phương trình $x^2+3x+8=(x+5)\sqrt{x^2+x+2}$ 21/08/2021 Bởi Eliza Giải phương trình $x^2+3x+8=(x+5)\sqrt{x^2+x+2}$
Đáp án: `S={1;-2;-7/5}` Giải thích các bước giải: `x^2+3x+8=(x+5)sqrt(x^2+x+2)` `=>(x^2+3x+8)^2=(x+5)^2.(x^2+x+2)` `=>(x^2+3x+8)(x^2+3x+8)=(x^2+10x+25)(x^2+x+2)` `=>x^2(x^2+3x+8)+3x(x^2+3x+8)+8(x^2+3x+8)=x^2(x^2+x+2)+10x(x^2+x+2)+25(x^2+x+2)` `=>x^4+3x^3+8x^2+3x^3+9x^2+24x+8x^2+24x+64=x^4+x^3+2x^2+10x^3+10x^2+20x+25x^2+25x+50` `=>x^4+6x^3+25x^2+48x+64=x^4+11x^3+37x^2+45x+50` `=>5x^3+12x^2-3x-14=0` `=>5x^3-5x^2+17x^2-17x+14x-14=0` `=>5x^2(x-1)+17x(x-1)+14(x-1)=0` `=>(x-1)(5x^2+17x+14)=0` `=>(x-1)(5x^2+10x+7x+14)=0` `=>(x-1)[5x(x+2)+7(x+2)]=0` `=>(x-1)(x+2)(5x+7)=0` `=>`\(\left[ \begin{array}{l}x-1=0\\x+2=0\\5x+7=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=-2\\5x=-7\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=-2\\x=-\dfrac{7}{5}\end{array} \right.\) Vậy `S={1;-2;-7/5}.` Bình luận
Đáp án:
`S={1;-2;-7/5}`
Giải thích các bước giải:
`x^2+3x+8=(x+5)sqrt(x^2+x+2)`
`=>(x^2+3x+8)^2=(x+5)^2.(x^2+x+2)`
`=>(x^2+3x+8)(x^2+3x+8)=(x^2+10x+25)(x^2+x+2)`
`=>x^2(x^2+3x+8)+3x(x^2+3x+8)+8(x^2+3x+8)=x^2(x^2+x+2)+10x(x^2+x+2)+25(x^2+x+2)`
`=>x^4+3x^3+8x^2+3x^3+9x^2+24x+8x^2+24x+64=x^4+x^3+2x^2+10x^3+10x^2+20x+25x^2+25x+50`
`=>x^4+6x^3+25x^2+48x+64=x^4+11x^3+37x^2+45x+50`
`=>5x^3+12x^2-3x-14=0`
`=>5x^3-5x^2+17x^2-17x+14x-14=0`
`=>5x^2(x-1)+17x(x-1)+14(x-1)=0`
`=>(x-1)(5x^2+17x+14)=0`
`=>(x-1)(5x^2+10x+7x+14)=0`
`=>(x-1)[5x(x+2)+7(x+2)]=0`
`=>(x-1)(x+2)(5x+7)=0`
`=>`\(\left[ \begin{array}{l}x-1=0\\x+2=0\\5x+7=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=-2\\5x=-7\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=-2\\x=-\dfrac{7}{5}\end{array} \right.\)
Vậy `S={1;-2;-7/5}.`