Question

Решите пожалуйста, 9 и 10​​

Answer 1

Объяснение:

9.

$(\frac{4a-1}{a^{2}+4a } +\frac{4a+1}{a^2-4a })*\frac{a^{2} -16}{a^{2} +1}= (\frac{4a-1}{a*(a+4) } +\frac{4a+1}{a*(a-4a) })*\frac{a^{2} -16}{a^{2} +1}=\\=\frac{(4a-1)*(a-4)+(4a+1)*(a+4)}{a*(a+4)*(a-4)}*\frac{a^{2} -16}{a^{2} +1}=\frac{4a^{2}-17a+4+4a+17a+4 }{a*(a^{2}-16) } *\frac{a^{2} -16}{a^{2} +1}=$

$=\frac{8a^{2}+8 }{a*(a^{2} -16)} *\frac{a^{2} -16}{a^{2} +1}= =\frac{8*(a^{2}+1) }{a*(a^{2} -16)} *\frac{a^{2} -16}{a^{2} +1}=\frac{8*(a^{2}+1)*(a^{2}-16) }{a*(a^{2}-16)*(a^{2}+1) } =\\=\frac{8}{5^{-1}}=\frac{8}{\frac{1}{5} } =8*5=40.$

10.

$16x^4-40x^2+25-10*(4x^{2} -5)-11=0\\16x^4-40x^2+25-40x^{2} -50-11=0\\16x^4-80x^{2} +64=0\\(4x^{2} )^2-2*4x^{2} *10+100-36=0\\(4x^{2} )^2-2*4x^{2} *10+10^2-36=0\\((4x^{2} )^2-10)^2-6^2=0\\(4x^{2} -10+6)*(4x^{2}-10-6)=0\\ (4x^{2} -4)*(4x^{2} -16)=0\\4x^{2} -4=0|:4\\x^{2} -1=0\\x^{2}-1^2=0\\(x+1)*(x-1)=0\\x_1=-1;x_2=1.\\4x^{2}-16=0|:4\\x^{2} -4=0\\x^{2} -2^2=0\\(x+2)*(x-2)=0\\x_3=-2;x_4=2.$

Ответ: x₁=-1    x₂=1    x₃=-2    x₄=2.