Неравенство равносильно неравенству
![2^x*2^{\sqrt{x}}+(2^x)^2\leq6*(2^{\sqrt{x}})^2](https://tex.z-dn.net/?f=2%5Ex%2A2%5E%7B%5Csqrt%7Bx%7D%7D%2B%282%5Ex%29%5E2%5Cleq6%2A%282%5E%7B%5Csqrt%7Bx%7D%7D%29%5E2)
Пусть ![2^x=a, 2^{\sqrt{x}}=b, a>0, b>0](https://tex.z-dn.net/?f=2%5Ex%3Da%2C+2%5E%7B%5Csqrt%7Bx%7D%7D%3Db%2C+a%3E0%2C+b%3E0)
![ab+a^2\leq6b^2\\a^2+ab-6b^2\leq0\\a^2+3ab-2ab-6b^2\leq0\\a(a+3b)-2b(a+3b)\leq0\\(a+3b)(a-2b)\leq0](https://tex.z-dn.net/?f=ab%2Ba%5E2%5Cleq6b%5E2%5C%5Ca%5E2%2Bab-6b%5E2%5Cleq0%5C%5Ca%5E2%2B3ab-2ab-6b%5E2%5Cleq0%5C%5Ca%28a%2B3b%29-2b%28a%2B3b%29%5Cleq0%5C%5C%28a%2B3b%29%28a-2b%29%5Cleq0)
I случай:
![\begin{equation*}\begin{cases}a+3b\leq0\\a-2b\geq0 \end{cases}\end{equation*}\Rightarrow \begin{equation*}\begin{cases} a\leq-3b\\a\geq2b \end{cases}\end{equation*}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7Da%2B3b%5Cleq0%5C%5Ca-2b%5Cgeq0+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow+%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7D+a%5Cleq-3b%5C%5Ca%5Cgeq2b+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D)
Так как b > 0, -3b < 0 ⇒ a < 0, но a > 0 - противоречие, значит, неравенство не имеет решений, следовательно, и система тоже не имеет решений.
II случай:
![\begin{equation*}\begin{cases}a+3b\geq0\\a-2b\leq0 \end{cases}\end{equation*}\Rightarrow\begin{equation*}\begin{cases}a\geq-3b\\a\leq2b \end{cases}\end{equation*}\Rightarrow\begin{equation*}\begin{cases}\frac{a}{b}\geq-3\\\frac{a}{b}\leq2 \end{cases}\end{equation*}\Rightarrow\begin{equation*}\begin{cases}\frac{2^x}{2^{\sqrt{x}}}\geq-3\\\frac{2^x}{2^{\sqrt{x}}}\leq2 \end{cases}\end{equation*}\Rightarrow](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7Da%2B3b%5Cgeq0%5C%5Ca-2b%5Cleq0+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7Da%5Cgeq-3b%5C%5Ca%5Cleq2b+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7D%5Cfrac%7Ba%7D%7Bb%7D%5Cgeq-3%5C%5C%5Cfrac%7Ba%7D%7Bb%7D%5Cleq2+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7D%5Cfrac%7B2%5Ex%7D%7B2%5E%7B%5Csqrt%7Bx%7D%7D%7D%5Cgeq-3%5C%5C%5Cfrac%7B2%5Ex%7D%7B2%5E%7B%5Csqrt%7Bx%7D%7D%7D%5Cleq2+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow)
![\Rightarrow \begin{equation*}\begin{cases}2^{x-\sqrt{x}}\geq-3\\ 2^{x-\sqrt{x}}\leq2^1 \end{cases}\end{equation*}\Rightarrow x-\sqrt{x}\leq1 \Leftrightarrow \sqrt{x}\geq x-1](https://tex.z-dn.net/?f=%5CRightarrow+%5Cbegin%7Bequation%2A%7D%5Cbegin%7Bcases%7D2%5E%7Bx-%5Csqrt%7Bx%7D%7D%5Cgeq-3%5C%5C+2%5E%7Bx-%5Csqrt%7Bx%7D%7D%5Cleq2%5E1+%5Cend%7Bcases%7D%5Cend%7Bequation%2A%7D%5CRightarrow+x-%5Csqrt%7Bx%7D%5Cleq1+%5CLeftrightarrow+%5Csqrt%7Bx%7D%5Cgeq+x-1)
Если x < 1, то x ∈ [0; 1). Если x ≥ 1:
![x\geq x^2-2x+1\\x^2-3x+1\leq0\\x^2-3x+1=0\\x=\frac{3\pm\sqrt{5}}{2} \\ x\in[\frac{3-\sqrt{5}}{2}; \frac{3+\sqrt{5}}{2}]](https://tex.z-dn.net/?f=x%5Cgeq+x%5E2-2x%2B1%5C%5Cx%5E2-3x%2B1%5Cleq0%5C%5Cx%5E2-3x%2B1%3D0%5C%5Cx%3D%5Cfrac%7B3%5Cpm%5Csqrt%7B5%7D%7D%7B2%7D+%5C%5C+x%5Cin%5B%5Cfrac%7B3-%5Csqrt%7B5%7D%7D%7B2%7D%3B+%5Cfrac%7B3%2B%5Csqrt%7B5%7D%7D%7B2%7D%5D)
Так как
, решением данного случая будет ![[1; \frac{3+\sqrt{5}}{2}]](https://tex.z-dn.net/?f=%5B1%3B+%5Cfrac%7B3%2B%5Csqrt%7B5%7D%7D%7B2%7D%5D)
Объединим и получим ответ.
Ответ: ![[0; \frac{3+\sqrt{5}}{2}]](https://tex.z-dn.net/?f=%5B0%3B+%5Cfrac%7B3%2B%5Csqrt%7B5%7D%7D%7B2%7D%5D)
<span>Разложить на множители:
10ab+15b2= 5в(2а+3в) x2-10x+25= (х-5)(х-5)
27a2+18ab=9а(3а+2в) y2+6y+9=(у+3)(у+3)
x2+xy-3x-3y=(х-3)(х+у) (a+1)2-9a2=(1-2а)(4а+1)
2xy-5y2-6x+15y=(2х-5у)(у+3) b2-(b-2)2= 2(2в-2)
a4-16=(а-2)(а+2)(а+4) x3+8y3=(х+2у)(х2-2ху+у2)
49-b4=(7-в2)(7+в2) x3-27y3=(х-3у)(х2+3ху+у2)</span>
X+y=4
x=4-y
x-y=8
4-y-y=8
4-2y=8
2y=4-8
2y=-4
y=-2
x=4-(-2)
x=6
(6;-2)