Question

    Giải phương trình :a;  \(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)

                               b, \(x^4+x^3-4x^2+5x-3=0\)

                               c,\(x^4-10.2^x+16=0\)

Answer 1

a) \(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)

\(\Leftrightarrow\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

\(\Leftrightarrow x=123\)

c) \(x^4-10.2^x+16=0\)

\(\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Ta có: 

\(2^x=t\)

\(\Rightarrow t^2-10t+16=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)