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(2,5x - 4/3)2 - 5/6 = -7/18
=> (2,5x - 4/3)2 = -7/18 + 5/6
=> (2,5x - 4/3)2 = 4/9
=> (2,5x - 4/3)2 = (2/3)2
=> \(\orbr{\begin{cases}2,5x-\frac{4}{3}=\frac{2}{3}\\2,5x-\frac{4}{3}=-\frac{2}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)
\(\left(2,5x-\frac{4}{3}\right)^2-\frac{5}{6}=-\frac{7}{18}\)
\(\left(2,5x-\frac{4}{3}\right)^2=-\frac{7}{18}+\frac{5}{6}\)
\(\left(2,5x-\frac{4}{3}\right)^2=\frac{4}{9}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)
Bài 3:
a) Đặt f(x)=0
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) Đặt f(x)=0
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Bài 3:
c) Đặt f(x)=0
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
d) Đặt f(x)=0
\(\Leftrightarrow x^4+2=0\)
\(\Leftrightarrow x^4=-2\)(Vô lý)
\(M\left(x\right)=P\left(x\right)+Q\left(x\right)=2,5x^6-4+2,5x^5-6x^3+2x^2\)-5x+\(3x-2,5x^6-x^2+5-2,5x^5+6x^3\)
=\(\left(2,5x^6-2,5x^6\right)\)+\(\left(2,5x^5-2,5x^5\right)\)\(\left(-6x^3+6x^3\right)\)+\(\left(2x^2-x^2\right)\)+\(\left(-5x+3x\right)\)+(-4+5)
= \(x^2-2x+1\)
\(3,5x+\left(-1,5\right)x=-4,9-2,7\)
\(3,5x+\left(-1,5\right)x=-7,6\)
\(x.\left[3,5+\left(-1,5\right)\right]=-7,6\)
\(x.2\) \(=-7,6\)
\(x\) \(=-3,8\)