-------------------------------
![\sqrt{x-3}=3 \sqrt[4]{x-3}+4\\\\
\sqrt[4]{x-3}=t \geq 0\\\\
t^2=3t+4\\\\
t^2-3t-4=0\\\\
t^2+t-4t-4=0\\\\
t(t+1)-4(t+1)=0\\\\
(t-4)(t+1)=0\\\\
t-4=0\ \ or\ \ t+1=0\\\\
t=4\ \ or\ \ t=-1\\\\
\sqrt[4]{x-3}=4\\\\
\left \{ {{x-3=4^4} \atop {4 \geq 0}} \right. \\\\
x=4^4+3=256+3=259\\\\](https://tex.z-dn.net/?f=+%5Csqrt%7Bx-3%7D%3D3+%5Csqrt%5B4%5D%7Bx-3%7D%2B4%5C%5C%5C%5C%0A++%5Csqrt%5B4%5D%7Bx-3%7D%3Dt+%5Cgeq+0%5C%5C%5C%5C%0At%5E2%3D3t%2B4%5C%5C%5C%5C%0At%5E2-3t-4%3D0%5C%5C%5C%5C%0At%5E2%2Bt-4t-4%3D0%5C%5C%5C%5C%0At%28t%2B1%29-4%28t%2B1%29%3D0%5C%5C%5C%5C%0A%28t-4%29%28t%2B1%29%3D0%5C%5C%5C%5C%0At-4%3D0%5C+%5C+or%5C+%5C+t%2B1%3D0%5C%5C%5C%5C%0At%3D4%5C+%5C+or%5C+%5C+t%3D-1%5C%5C%5C%5C%0A%5Csqrt%5B4%5D%7Bx-3%7D%3D4%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx-3%3D4%5E4%7D+%5Catop+%7B4+%5Cgeq+0%7D%7D+%5Cright.+%5C%5C%5C%5C%0Ax%3D4%5E4%2B3%3D256%2B3%3D259%5C%5C%5C%5C "\sqrt{x-3}=3 \sqrt[4]{x-3}+4\\\\
\sqrt[4]{x-3}=t \geq 0\\\\
t^2=3t+4\\\\
t^2-3t-4=0\\\\
t^2+t-4t-4=0\\\\
t(t+1)-4(t+1)=0\\\\
(t-4)(t+1)=0\\\\
t-4=0\ \ or\ \ t+1=0\\\\
t=4\ \ or\ \ t=-1\\\\
\sqrt[4]{x-3}=4\\\\
\left \{ {{x-3=4^4} \atop {4 \geq 0}} \right. \\\\
x=4^4+3=256+3=259\\\\")
-------------------------------
-----------------------------
![x^2-2x-3=x^2-2x+1-4=(x-1)^2-2^2=\\\\=[(x-1)-2]*[(x-1)+2]=(x-3)(x+1)](https://tex.z-dn.net/?f=x%5E2-2x-3%3Dx%5E2-2x%2B1-4%3D%28x-1%29%5E2-2%5E2%3D%5C%5C%5C%5C%3D%5B%28x-1%29-2%5D%2A%5B%28x-1%29%2B2%5D%3D%28x-3%29%28x%2B1%29 "x^2-2x-3=x^2-2x+1-4=(x-1)^2-2^2=\\\\=[(x-1)-2]*[(x-1)+2]=(x-3)(x+1)")
![(0.4^{\frac{1}{x^2-2x-3}})^{6-x} \geq 1\\\\
0.4^{\frac{6-x}{(x-3)(x+1)}} \geq 0.4^0\\\\
0.4^{-\frac{(x-6)}{(x-3)(x+1)}} \geq 0.4^0\\\\
-\frac{(x-6)}{(x-3)(x+1)} \leq 0\\\\
\frac{x-6}{(x-3)(x+1)} \geq 0\\\\
-----(-1)++++++(3)-------[6]+++++\ \textgreater \ x\\\\
x\in(-1;\ 3)\cup[6;\ +\infty)](https://tex.z-dn.net/?f=%280.4%5E%7B%5Cfrac%7B1%7D%7Bx%5E2-2x-3%7D%7D%29%5E%7B6-x%7D+%5Cgeq+1%5C%5C%5C%5C%0A0.4%5E%7B%5Cfrac%7B6-x%7D%7B%28x-3%29%28x%2B1%29%7D%7D+%5Cgeq+0.4%5E0%5C%5C%5C%5C%0A0.4%5E%7B-%5Cfrac%7B%28x-6%29%7D%7B%28x-3%29%28x%2B1%29%7D%7D+%5Cgeq+0.4%5E0%5C%5C%5C%5C%0A-%5Cfrac%7B%28x-6%29%7D%7B%28x-3%29%28x%2B1%29%7D+%5Cleq+0%5C%5C%5C%5C%0A%5Cfrac%7Bx-6%7D%7B%28x-3%29%28x%2B1%29%7D+%5Cgeq+0%5C%5C%5C%5C%0A-----%28-1%29%2B%2B%2B%2B%2B%2B%283%29-------%5B6%5D%2B%2B%2B%2B%2B%5C+%5Ctextgreater+%5C+x%5C%5C%5C%5C%0Ax%5Cin%28-1%3B%5C+3%29%5Ccup%5B6%3B%5C+%2B%5Cinfty%29 "(0.4^{\frac{1}{x^2-2x-3}})^{6-x} \geq 1\\\\
0.4^{\frac{6-x}{(x-3)(x+1)}} \geq 0.4^0\\\\
0.4^{-\frac{(x-6)}{(x-3)(x+1)}} \geq 0.4^0\\\\
-\frac{(x-6)}{(x-3)(x+1)} \leq 0\\\\
\frac{x-6}{(x-3)(x+1)} \geq 0\\\\
-----(-1)++++++(3)-------[6]+++++\ \textgreater \ x\\\\
x\in(-1;\ 3)\cup[6;\ +\infty)")
---------------------------
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Если в уравнении положить
, то из уравнения следует, что и
, чего быть не может (синус и косинус того же аргумента не могут равняться нулю одновременно, это противоречит основному тригонометрическому тождеству), т.е. в данном уравнении
Это означает, что мы можем делить уравнение на
и решение нового уравнения совпадать с решениям исходного (равносильный переход)
