1)(х^2+4)^2+(х^2+4)-30=02)(х^2-8)^2+3,5(х^2-8)-2=03)(1-х^2)^2+3,7(1-х^2)+2,1=04)(1+х^2)^2+0,5(1+х^2)-5=0. Created by 787997. algebra-ru

1)(х^2+4)^2+(х^2+4)-30=02)(х^2-8)^2+3,5(х^2-8)-2=03)(1-х^2)^2+3,7(1-х^2)+2,1=04)...

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1)(х^2+4)^2+(х^2+4)-30=0
2)(х^2-8)^2+3,5(х^2-8)-2=0
3)(1-х^2)^2+3,7(1-х^2)+2,1=0
4)(1+х^2)^2+0,5(1+х^2)-5=0

Answers & Comments

skvrttt 1.
$(x^2+4)^2+x^2-26=0\\x^2+4=a\to a^2+a-30=0\\D=1+120=11^2\\\left[\begin{array}{ccc}x^2+4=a_1=\frac{-1+11}{2}=5\\x^2+4=a_2=\frac{-1-11}{2}=-6\end{array}\right\left[\begin{array}{ccc}x^2=1\\x^2=-10\end{array}\right\\x_{1,2}=б1$

2.
$(x^2-8)^2+3,5(x^2-8)-2=0\\x^2-8=a\to 2a^2+7a-4=0\\D=49+32=9^2\\\left[\begin{array}{ccc}x^2-8=a_1=\frac{-7+9}{4}=0,5\\x^2-8=a_2=\frac{-7-9}{4}=-4\end{array}\right\left[\begin{array}{ccc}x^2=7,5\\x^2=4\end{array}\right\\x_{1,2}=б\sqrt{7,5}=б5\sqrt{0,3}\\x_{3,4}=б2$

3.
$(1-x^2)^2+3,7(1-x^2)+2,1=0\\1-x^2=a\to 10a^2+37a+21=0\\D=1369-840=23^2\\\left[\begin{array}{ccc}1-x^2=a_1=\frac{-37+23}{20}=-0,7\\1-x^2=a_2=\frac{-37-23}{20}=-3\end{array}\right\left[\begin{array}{ccc}x^2=1,7\\x^2=4\end{array}\right\\x_{1,2}=б\sqrt{1,7}\\x_{3,4}=б2$

4.
$(1+x^2)^2+0,5(1+x^2)-5=0\\1+x^2=a\to 2a^2+a-10=0\\D=1+80=9^2\\\left[\begin{array}{ccc}1+x^2=a_1=\frac{-1+9}{4}=2\\1+x^2=a_2=\frac{-1-9}{4}=-2,5\end{array}\right\left[\begin{array}{ccc}x^2=1\\x^2=-3,5\end{array}\right\\x_{1,2}=б1$

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