1.решить уравнение: с проверкой! а)9^x-10∙3^x+9=0

To solve the equation $9^x-10\cdot 3^x+9=0$, we will use the method of substitution. 1. Simplify the Equation First, observe that $9^x$ can be written as $(3^2{)}^x$, which is equal to $(3^x{)}^2$. The equation becomes: $(3^x{)}^2-10\cdot 3^x+9=0$2. Substitution Let $t=3^x$. Since any exponential function with a positive base is always positive, we must have $t>0$. The equation now turns into a standard quadratic: $t^2-10t+9=0$3. Solve the Quadratic Equation We can solve this using the quadratic formula or by factoring. Looking for two numbers that multiply to $9$ and add up to $-10$, we find $-1$ and $-9$: $(t-1)(t-9)=0$This gives us two possible values for $t$:

1. $t_1=1$ $t_2=9$

Both values are positive, so they are both valid for the next step. 4. Back-Substitution (Finding $x$) Now we substitute $3^x$ back in for $t$: Case 1: $3^x=1$ Since $3^0=1$, we have: $x_1=0$ Case 2: $3^x=9$ Since $9=3^2$, we have: $x_2=2$ 5. Verification (Check) We test both solutions in the original equation $9^x-10\cdot 3^x+9=0$. For $x=0$: $9^0-10\cdot 3^0+9=1-10\cdot 1+9$ $1-10+9=0$ $0=0$ (Correct) For $x=2$: $9^2-10\cdot 3^2+9=81-10\cdot 9+9$ $81-90+9=-9+9$ $0=0$ (Correct) Final Answer: $x\in \{0,2\}$ Would you like me to solve another equation from your list using this same step-by-step method?