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a. Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|x-1|+|x-4|=|x-1|+|4-x|\geq |x-1+4-x|=3$
$|x-2|+|x-4|=|x-2|+|4-x|\geq |x-2+4-x|=2$
$|x-4|\geq 0$
Cộng theo vế:
$A\geq 5$
Vậy $A_{\min}=5$. Giá trị này đạt tại \(\left\{\begin{matrix} (x-1)(4-x)\geq 0\\ (x-2)(4-x)\geq 0\\ x-4=0\end{matrix}\right.\Leftrightarrow x=4\)
c. Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ thì:
$|x-3|+|x-8|=|x-3|+|8-x|\geq |x-3+8-x|=5$
$|x-5|+|x-8|=|x-5|+|8-x|\geq |x-5+8-x|=3$
$3|x-8|\geq 0$
Cộng theo vế:
$C\geq 8$. Vậy $C_{\min}=8$. Giá trị này đạt tại $x=8$
Bài 10:
\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{\left(2^2\right)^6\cdot3^6+\left(2^3\right)^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\\ =\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}\\ =\dfrac{3-1}{3\cdot\left(3+1\right)}\\ =\dfrac{2}{3\cdot4}\\ =\dfrac{1}{6}\)
\(B=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\\ =\dfrac{2\cdot3\cdot5\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot\left(3^2\right)^7-2^2\cdot3\cdot\left(2^3\right)^5\cdot7^5\cdot}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{17}\cdot3^{14}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{31}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{5\cdot\left(3^{30}\cdot2^{15}-2^{17}\cdot7^5\right)}{3\cdot\left(2^{15}\cdot3^{30}-2^{17}\cdot7^5\right)}\\ =\dfrac{5}{3}\)
Bài 8:
\(\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y+z=\dfrac{1}{3}\\x+z=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{3}-z+\dfrac{1}{6}-z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-2z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2z=0\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=0\\y=\dfrac{1}{3}-0=\dfrac{1}{3}\\x=\dfrac{1}{6}-0=\dfrac{1}{6}\end{matrix}\right.\)
Bài 5:
\(A=\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
\(=\dfrac{15-4+1}{10}:\dfrac{9-4+1}{12}\)
\(=\dfrac{12}{10}\cdot\dfrac{12}{6}=\dfrac{6}{5}\cdot2=\dfrac{12}{5}\)