Решениеееееееееееееееееее
ОДЗ
х≥0
![2(lg2-1)+lg( 5^{ \sqrt{x} } +1) \leq lg( 5^{1- \sqrt{x} } +5) \\ \\ 2(lg2-lg10)+lg( 5^{ \sqrt{x} } +1) \leq lg( 5^{1- \sqrt{x} } +5) \\ \\ 2(lg \frac{2}{10} )+lg( 5^{ \sqrt{x} } +1) \leq lg( 5^{1- \sqrt{x} } +5)](https://tex.z-dn.net/?f=2%28lg2-1%29%2Blg%28++5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%29+%5Cleq+lg%28+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%29+%5C%5C++%5C%5C+2%28lg2-lg10%29%2Blg%28++5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%29+%5Cleq+lg%28+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%29+%5C%5C++%5C%5C+2%28lg+%5Cfrac%7B2%7D%7B10%7D+%29%2Blg%28++5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%29+%5Cleq+lg%28+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%29+)
![lg (\frac{1}{5})^2+lg( 5^{ \sqrt{x} } +1) \leq lg( 5^{1- \sqrt{x} } +5) \\ \\lg (\frac{1}{5})^2\cdot( 5^{ \sqrt{x} } +1) \leq lg( 5^{1- \sqrt{x} } +5) \\ \\ \frac{5^{ \sqrt{x} } +1}{25} \leq 5^{1- \sqrt{x} } +5](https://tex.z-dn.net/?f=lg+%28%5Cfrac%7B1%7D%7B5%7D%29%5E2%2Blg%28++5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%29+%5Cleq+lg%28+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%29+%5C%5C++%5C%5Clg+%28%5Cfrac%7B1%7D%7B5%7D%29%5E2%5Ccdot%28++5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%29+%5Cleq+lg%28+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%29+%5C%5C++%5C%5C+%5Cfrac%7B5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1%7D%7B25%7D++%5Cleq+5%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5+)
![5^{ \sqrt{x} } +1 \leq (5^{1- \sqrt{x} } +5})\cdot 25 \\ \\ 5^{ \sqrt{x} } +1 \leq125\cdot 5^{- \sqrt{x} } +125 \\ \\ (5^{ \sqrt{x} })^{2} -124\cdot 5^{ \sqrt{x} } -125 \leq 0](https://tex.z-dn.net/?f=5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1++%5Cleq+%285%5E%7B1-+%5Csqrt%7Bx%7D+%7D+%2B5%7D%29%5Ccdot+25++%5C%5C++%5C%5C+5%5E%7B+%5Csqrt%7Bx%7D+%7D+%2B1++%5Cleq125%5Ccdot+5%5E%7B-+%5Csqrt%7Bx%7D+%7D+%2B125++%5C%5C++%5C%5C+%285%5E%7B+%5Csqrt%7Bx%7D+%7D%29%5E%7B2%7D+-124%5Ccdot+5%5E%7B+%5Csqrt%7Bx%7D+%7D+-125+%5Cleq+0++)
Замена переменной
![5^{ \sqrt{x} }=t \\ \\ (5^{ \sqrt{x} })^2=t^2](https://tex.z-dn.net/?f=5%5E%7B+%5Csqrt%7Bx%7D+%7D%3Dt+%5C%5C++%5C%5C+%285%5E%7B+%5Csqrt%7Bx%7D+%7D%29%5E2%3Dt%5E2+)
Так как показательная функция принимает только положительные значения, то t >0
t² - 124 t - 125 ≤ 0 (*)
D = (-124)²-4·(-125)=4·(4·31²+125)=4·(3844+125)=4·3969=(2·63)²=126²
t₁=(124-126)/2=-1 или t₂=(124+126)/2=125
Решение неравенства (*)
-1≤ t≤125
Но с учетом условия t >0, получим ответ
0 < t ≤ 125
t > 0 при любом х из ОДЗ : х≥0
![5^{ \sqrt{x} } \leq 125 \\ \\ 5^{ \sqrt{x} } \leq 5^3 \\ \\ \sqrt{x} \leq 3 \\ 0 \leq x \leq 9](https://tex.z-dn.net/?f=5%5E%7B+%5Csqrt%7Bx%7D+%7D+%5Cleq+125+%5C%5C++%5C%5C+5%5E%7B+%5Csqrt%7Bx%7D+%7D+%5Cleq+5%5E3+%5C%5C++%5C%5C++%5Csqrt%7Bx%7D++%5Cleq+3+%5C%5C+0+%5Cleq+x+%5Cleq+9)
Q= 15/30=1/2
b4=b1*q^3=30*1/8=15/4
S5=30/1-1/2=60