a) Bạn adct \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

Ta cóA= \(\left|x-7\right|+\left|x+5\right|=\left|7-x\right|+\left|x+5\right|\ge\left|7-x+x+5\right|\)

=> \(\left|7-x\right|+\left|x+5\right|\ge12\) vậy minA=12

b)Ta có \(\left(2x-1\right)^2-3\left|2x-1\right|+2=\left|2x-1\right|^2-2\left|2x-1\right|.\frac{3}{2}+\frac{9}{4}-\frac{1}{4}=\left(\left|2x-1\right|-\frac{3}{2}\right)^2-\frac{1}{4}\)=>minA=-1/4

còn câu c thì bạn làm giống câu a nhé

Có: \(\left|x+5\right|\ge x+5;\left|3-x\right|\ge3-x\forall x\)

\(\Rightarrow A=\left|x+5\right|+\left|3-x\right|\ge x+5+3-x=8\)

Dấu ''='' xảy ra khi \(\begin{cases}x+5\ge0\\3-x\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge-5\\x\le3\end{cases}\)

Vậy GTNN của A là 8 khi \(-5\le x\le3\)

\(\left(x+4\right).11=x.14\\ \Leftrightarrow11x+44=14x\\ \Leftrightarrow11x-14x=-44\\ \Leftrightarrow-3x=-44\\ \Leftrightarrow x=\dfrac{44}{3}\)

Vậy \(x=\dfrac{44}{3}\)

\(\left(x+4\right).11=x.14\)

\(11x+44-14x=0\)

\(-3x-44=0\)

\(-3x=44\)

\(x=\dfrac{44}{3}\)

(x + 4).11 = x.14

11.(x + 4) = 14x

11x = 14x - 44

11x - 14x = 14x - 44 + 14x

-3x = -44

x = 44/3

a) Rút gọn :

ĐKXĐ : \(x\ne4,x\ne3\)

Ta có : \(Q=\frac{12x-45}{x^2-7x+12}-\frac{x+5}{x-4}+\frac{2x-3}{3-x}\)

\(=\frac{3\left(4x-15\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(x+5\right)\left(x-3\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(2x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)

\(=\frac{12x-45-x^2-2x+15-2x^2+11x-12}{\left(x-4\right)\left(x-3\right)}\)

\(=\frac{-3x^2+21x-42}{\left(x-4\right)\left(x-3\right)}\)

... Chắc tui rút gọn sai òi :))