Resolva as eguações irracionais :
a) √x+2 = 7
b) √5x-10 = √3x+2
c)3√x+8 = 20
d) ³√x²-x-4 =0
Soluções para a tarefa
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2
a)
![\sqrt{x+2}=7 \\ (\sqrt{x+2})^2=(7)^2 \\ x+2=49 \\ x=47 \\ \sqrt{x+2}=7 \\ (\sqrt{x+2})^2=(7)^2 \\ x+2=49 \\ x=47 \\](https://tex.z-dn.net/?f=+%5Csqrt%7Bx%2B2%7D%3D7+%5C%5C++%28%5Csqrt%7Bx%2B2%7D%29%5E2%3D%287%29%5E2+%5C%5C+x%2B2%3D49+%5C%5C+x%3D47+%5C%5C++)
b)
![\sqrt{5x-10} = \sqrt{3x+2 \\ } \\ (\sqrt{5x-10})^2 =( \sqrt{3x+2 \\ })^2 \\ 5x-10=3x+2 \\ 2x=12 \\ x=6 \sqrt{5x-10} = \sqrt{3x+2 \\ } \\ (\sqrt{5x-10})^2 =( \sqrt{3x+2 \\ })^2 \\ 5x-10=3x+2 \\ 2x=12 \\ x=6](https://tex.z-dn.net/?f=+%5Csqrt%7B5x-10%7D++%3D+%5Csqrt%7B3x%2B2+%5C%5C+%7D++%5C%5C+%28%5Csqrt%7B5x-10%7D%29%5E2++%3D%28+%5Csqrt%7B3x%2B2+%5C%5C+%7D%29%5E2+%5C%5C+5x-10%3D3x%2B2+%5C%5C+2x%3D12+%5C%5C+x%3D6+)
c)
![3 \sqrt{ x+8}=20 \\ (3 \sqrt{ x+8})^2=(20)^2 \\ 9(x+8)=400 \\ 9x+72=400 \\ x= \frac{400-72}{9} = \frac{328}{9} 3 \sqrt{ x+8}=20 \\ (3 \sqrt{ x+8})^2=(20)^2 \\ 9(x+8)=400 \\ 9x+72=400 \\ x= \frac{400-72}{9} = \frac{328}{9}](https://tex.z-dn.net/?f=3+%5Csqrt%7B+x%2B8%7D%3D20+%5C%5C+%283+%5Csqrt%7B+x%2B8%7D%29%5E2%3D%2820%29%5E2++%5C%5C+9%28x%2B8%29%3D400+%5C%5C+9x%2B72%3D400+%5C%5C+x%3D+%5Cfrac%7B400-72%7D%7B9%7D+%3D+%5Cfrac%7B328%7D%7B9%7D++)
d)
![\sqrt[3]{ x^{2}-x-4 }=0 \\ (\sqrt[3]{ x^{2}-x-4 })^3=(0)^3 \\ x^{2}-x-4=0 \\ x'= \frac{1+ \sqrt{17} }{2} \\ x"= \frac{1-\sqrt{17} }{2} \sqrt[3]{ x^{2}-x-4 }=0 \\ (\sqrt[3]{ x^{2}-x-4 })^3=(0)^3 \\ x^{2}-x-4=0 \\ x'= \frac{1+ \sqrt{17} }{2} \\ x"= \frac{1-\sqrt{17} }{2}](https://tex.z-dn.net/?f=+%5Csqrt%5B3%5D%7B+x%5E%7B2%7D-x-4+%7D%3D0+%5C%5C+%28%5Csqrt%5B3%5D%7B+x%5E%7B2%7D-x-4+%7D%29%5E3%3D%280%29%5E3+%5C%5C++x%5E%7B2%7D-x-4%3D0+%5C%5C++x%27%3D+%5Cfrac%7B1%2B+%5Csqrt%7B17%7D+%7D%7B2%7D+++%5C%5C+x%22%3D+%5Cfrac%7B1-%5Csqrt%7B17%7D+%7D%7B2%7D++)
b)
c)
d)
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