Đặt BĐT cần c/m là A

Dự đoán đẳng thức xảy ra khi a = b = c

Áp dụng BĐT Cauchy cho 3 số không âm:

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\)

\(\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(a+c\right)}.\frac{a+b}{8}.\frac{a+c}{8}}=\frac{3a}{4}\)

\(\frac{b^3}{\left(b+c\right)\left(b+a\right)}+\frac{b+c}{8}+\frac{b+a}{8}\)

\(\ge3\sqrt[3]{\frac{b^3}{\left(b+c\right)\left(b+a\right)}.\frac{b+c}{8}.\frac{b+a}{8}}=\frac{3b}{4}\)

\(\frac{c^3}{\left(c+a\right)\left(c+b\right)}+\frac{c+a}{8}+\frac{c+b}{8}\)

\(\ge3\sqrt[3]{\frac{c^3}{\left(c+a\right)\left(c+b\right)}.\frac{c+a}{8}.\frac{c+b}{8}}=\frac{3c}{4}\)

Cộng từng vế của các BĐT trên, ta được:

\(A+\frac{2\left(a+b+c\right)}{4}\ge\frac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow A\ge\frac{3}{4}\)

(Dấu "="\(\Leftrightarrow a=b=c\))

\(|x-3|-12=|-5|\)

\(\Leftrightarrow|x-3|-12=5\)

\(\Leftrightarrow|x-3|=17\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=17\\x-3=-17\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=20\\x=-17\end{cases}}\)

Vậy \(x=\left\{20;-17\right\}\)

\(|x-3|-12=|-5|\)\(\Leftrightarrow|x-3|-12=5\)

\(\Leftrightarrow|x-3|-12=5\)\(\Leftrightarrow|x-3|=17\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=-17\\x-3=17\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-14\\x=20\end{cases}}\)

Vậy \(x=-14\)hoặc \(x=20\)

a) \(x^2+x+1=x\left(x+1\right)+1\)

Vì \(x\inℤ\)\(\Rightarrow x\left(x+1\right)⋮x+1\)\(\Rightarrow\)Để \(x^2+x+1⋮x+1\)thì \(1⋮x+1\)

\(\Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{-2;0\right\}\)

Vậy \(x\in\left\{-2;0\right\}\)

b) \(3x-8=3x-12+4=3\left(x-4\right)+4\)

Vì \(3\left(x-4\right)⋮x-4\)\(\Rightarrow\)Để \(3x-8⋮x-4\)thì \(4⋮x-4\)

\(\Rightarrow x-4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Lập bảng giá trị ta có: 

Vậy \(x\in\left\{0;2;3;5;6;8\right\}\)

\(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Leftrightarrow3x-6+2x-2=10\)

\(\Leftrightarrow5x-8=10\)

\(\Leftrightarrow5x=18\)

\(\Leftrightarrow x=\frac{18}{5}\)

Vậy ..........

\(\left(x-1\right)^3=\left(x-1\right)\)\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

TH1: \(x-1=0\)\(\Leftrightarrow x=1\)

TH2: \(\left(x-1\right)^2-1=0\)\(\Leftrightarrow\left(x-1\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=-1\\x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy \(x\in\left\{0;1;2\right\}\)