A raiz da equação √x+4 + √x-4 / √x+4 - √x+4 =2 pertence ao intervalo
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Vamos substituir:
![a= \sqrt{x+4} \rightarrow a^2=x+4\\
b= \sqrt{x-4} \rightarrow b^2=x-4\\
\\
\frac{a+b}{a-b}=2 \\
\\
a+b=2(a-b)\\
(a+b)^2=[2(a-b)]^2\\
a^2+2ab+b^2=4(a^2-2ab+b^2)\\
a^2+2ab+b^2=4a^2-8ab+4b^2\\
3a^2-10ab+3b^2=0\\
3(x+4)-10 \sqrt{x^2-16}+3(x-4)=0 \\
3x+12-10\sqrt{x^2-16}+3x-12=0\\
6x=10\sqrt{x^2-16}\\
36x^2=100(x^2-16)\\
36x^2=100x^2-1600\\
64x^2=1600\\
x^2=25\\
\boxed{x=5}](https://tex.z-dn.net/?f=a%3D+%5Csqrt%7Bx%2B4%7D+%5Crightarrow+a%5E2%3Dx%2B4%5C%5C%0Ab%3D+%5Csqrt%7Bx-4%7D+%5Crightarrow+b%5E2%3Dx-4%5C%5C+%0A%5C%5C%0A%5Cfrac%7Ba%2Bb%7D%7Ba-b%7D%3D2+%5C%5C%0A%5C%5C%0Aa%2Bb%3D2%28a-b%29%5C%5C%0A%28a%2Bb%29%5E2%3D%5B2%28a-b%29%5D%5E2%5C%5C%0Aa%5E2%2B2ab%2Bb%5E2%3D4%28a%5E2-2ab%2Bb%5E2%29%5C%5C%0Aa%5E2%2B2ab%2Bb%5E2%3D4a%5E2-8ab%2B4b%5E2%5C%5C%0A3a%5E2-10ab%2B3b%5E2%3D0%5C%5C%0A%0A3%28x%2B4%29-10+%5Csqrt%7Bx%5E2-16%7D%2B3%28x-4%29%3D0+%5C%5C%0A3x%2B12-10%5Csqrt%7Bx%5E2-16%7D%2B3x-12%3D0%5C%5C%0A6x%3D10%5Csqrt%7Bx%5E2-16%7D%5C%5C%0A36x%5E2%3D100%28x%5E2-16%29%5C%5C%0A36x%5E2%3D100x%5E2-1600%5C%5C%0A64x%5E2%3D1600%5C%5C%0Ax%5E2%3D25%5C%5C%0A%5Cboxed%7Bx%3D5%7D%0A%0A "a= \sqrt{x+4} \rightarrow a^2=x+4\\
b= \sqrt{x-4} \rightarrow b^2=x-4\\
\\
\frac{a+b}{a-b}=2 \\
\\
a+b=2(a-b)\\
(a+b)^2=[2(a-b)]^2\\
a^2+2ab+b^2=4(a^2-2ab+b^2)\\
a^2+2ab+b^2=4a^2-8ab+4b^2\\
3a^2-10ab+3b^2=0\\
3(x+4)-10 \sqrt{x^2-16}+3(x-4)=0 \\
3x+12-10\sqrt{x^2-16}+3x-12=0\\
6x=10\sqrt{x^2-16}\\
36x^2=100(x^2-16)\\
36x^2=100x^2-1600\\
64x^2=1600\\
x^2=25\\
\boxed{x=5}")
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