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Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a) Ta có: \(\left(3x-1\right)^2+\left(4x+5\right)^2=\left(5x-7\right)^2\)
\(\Leftrightarrow9x^2-6x+1+16x^2+40x+25=25x^2-70x+49\)
\(\Leftrightarrow25x^2+34x+26-25x^2+70x-49=0\)
\(\Leftrightarrow104x-23=0\)
\(\Leftrightarrow104x=23\)
hay \(x=\frac{23}{104}\)
Vậy: \(S=\left\{\frac{23}{104}\right\}\)
b) Ta có: \(\left(x-2\right)^3+\left(x+2\right)^3=2\left(x-3\right)\left(x^2+3x+9\right)\)
\(\Leftrightarrow\left(x-2+x+2\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\right]=2\left(x^3-27\right)\)
\(\Leftrightarrow2x\cdot\left(x^2-4x+4-x^2+4+x^2+4x+4\right)=2x^3-54\)
\(\Leftrightarrow2x\cdot\left(x^2+12\right)-2x^3+54=0\)
\(\Leftrightarrow2x^3+24x-2x^3+54=0\)
\(\Leftrightarrow24x=54\)
hay \(x=\frac{9}{4}\)
Vậy: \(S=\left\{\frac{9}{4}\right\}\)
c) Ta có: \(2014x-10.07=20.14x-1007\)
\(\Leftrightarrow2014x-10.07-20.14x+1007=0\)
\(\Leftrightarrow1993.86x+1017.07=0\)
\(\Leftrightarrow1993.86x=-1017.07\)
\(\Leftrightarrow x=-\frac{101}{198}\)
Vậy: \(S=\left\{-\frac{101}{198}\right\}\)
d) Ta có: \(\frac{x-5}{2}+\frac{x-5}{3}-\frac{1}{4}=\frac{1}{2}+\frac{1}{3}-\frac{x-5}{4}\)
\(\Leftrightarrow\frac{x-5}{2}+\frac{x-5}{3}+\frac{x-5}{4}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Leftrightarrow x-5=1\)
hay x=6
Vậy: S={6}
Bài 2:
a)Ta thấy: \(\left|\frac{1}{2}-x\right|\ge0\)
\(\Rightarrow A\ge0\)
Dấu "=" xảy ra khi \(x=\frac{1}{2}\)
Vậy \(Min_A=0\) khi \(x=\frac{1}{2}\)
b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|\frac{1}{2}-x\right|+\left|x+2\right|\ge\left|\frac{1}{2}-x+x+2\right|=\frac{5}{2}\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_B=\frac{5}{2}\) khi x=0