1 ответ:
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1)
2)
ОДЗ: x+1>0
x> -1
![log_{4}(2y^2-2x+4)=6^{log_{6}(3lg \sqrt[3]{10} )}+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=3lg \sqrt[3]{10}+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=lg( \sqrt[3]{10} )^3+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=lg10+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=1+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=log_{4}4+log_{4}(x+1) \\ \\ log_{4}(2y^2-2x+4)=log_{4}(4(x+1))](https://tex.z-dn.net/?f=log_%7B4%7D%282y%5E2-2x%2B4%29%3D6%5E%7Blog_%7B6%7D%283lg+%5Csqrt%5B3%5D%7B10%7D++%29%7D%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3D3lg+%5Csqrt%5B3%5D%7B10%7D%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3Dlg%28+%5Csqrt%5B3%5D%7B10%7D+%29%5E3%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3Dlg10%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3D1%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3Dlog_%7B4%7D4%2Blog_%7B4%7D%28x%2B1%29+%5C%5C++%5C%5C+%0Alog_%7B4%7D%282y%5E2-2x%2B4%29%3Dlog_%7B4%7D%284%28x%2B1%29%29+)
2y²-2x+4=4(x+1)
2y²-2x-4x=4-4
2y²-6x=0
y²-3x=0
y²=3x
{y²=x+2
{y²=3x
x+2=3x
x-3x=-2
-2x=-2
x=1
y²=3*1
y²=3
y₁=√3
y₂= -√3
Ответ: (1; -√3);
(1; √3).
2)
ОДЗ: x+1>0
x> -1
2y²-2x+4=4(x+1)
2y²-2x-4x=4-4
2y²-6x=0
y²-3x=0
y²=3x
{y²=x+2
{y²=3x
x+2=3x
x-3x=-2
-2x=-2
x=1
y²=3*1
y²=3
y₁=√3
y₂= -√3
Ответ: (1; -√3);
(1; √3).