Ta có:\(\left|5x-3\right|=\left[{}\begin{matrix}5x-3\left(x\ge0\right)\\-\left(5x-3\right)=3-5x\left(x< 0\right)\end{matrix}\right.\)

Do đó, ta có 2 TH:

TH1:

\(5x-3-x\ge7\left(x\ge0\right)\\ \Leftrightarrow4x\ge7+3\\ \Leftrightarrow4x\ge10\\ \Leftrightarrow x\ge2,5\left(t/m\right)\)

TH2:

\(3-5x-x\ge7\left(x< 0\right)\\ \Leftrightarrow-6x\ge7-3\\ \Leftrightarrow-6x\ge4\\ \Leftrightarrow x\le-\dfrac{2}{3}\left(t/m\right)\)

Vậy \(x\ge2,5\) hoặc \(x\le-\dfrac{2}{3}\)

\(VT=27x^2-36x+12+\frac{8x}{y}\)

\(=\frac{8x}{1-x}+18x\left(1-x\right)+45x^2-54x+12\)

\(\ge45x^2-54x+12+24x\)

\(=45x^2-30x+12=5\left(9x^2-6x+\frac{12}{5}\right)\)

\(=5\left[\left(3x-1\right)^2+\frac{7}{5}\right]\ge7\)

Dấu = khi \(x=\frac{1}{3};y=\frac{2}{3}\)

\(2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2\left(a^2+b^2+c^2\right)+4\frac{ab+bc+ca}{abc}.\)

\(=2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)(vì abc=1)

\(=2\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\)

\(=2\left(a+b+c\right)^2\)

Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)(bất đẳng thức cô si cho ba số không âm)

Đặt \(a+b+c=x\ge3\)

Dễ thấy : \(2x^2-7x+3=\left(2x-1\right)\left(x-3\right)\ge0\)

Hay \(2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)

Dấu '=' xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=3\end{cases}\Leftrightarrow}a=b=c=1\)

Đặt A = a + b + c . 

Áp dụng BĐT Cosi cho 3 số thực dương ta có : \(A\ge3^3\sqrt{abc}=3\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-7\left(a+b+c\right)+3\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\cdot\frac{ab+bc+ca}{abc}-7\left(a+b+c\right)+3\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)-7\left(a+b+c\right)+3\)

\(\Leftrightarrow2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\)

\(\Leftrightarrow2A^2-7A+3=\left(2A-1\right)\left(A-3\right)\ge0\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(VT=27x^2-36x+12+\frac{15x-7}{1-x}+7\)

\(VT=\frac{-27x^3+63x^2-33x+5}{1-x}+7=\frac{\left(3x-1\right)^2\left(5-3x\right)}{1-x}+7\)

Do \(x< 1\Rightarrow\left\{{}\begin{matrix}5-3x>0\\1-x>0\end{matrix}\right.\) \(\Rightarrow\frac{\left(3x-1\right)^2\left(5-3x\right)}{1-x}\ge0\)

\(\Rightarrow VT\ge7\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{2}{3}\end{matrix}\right.\)

Ta có :

\(\left|3-5x\right|\ge7\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3-5x\ge7\\5x-3\ge7\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}-5x\ge4\\5x\ge10\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x\le-\frac{4}{5}\\x\ge2\end{array}\right.\)

Vậy ........

|3-5x|>=7

=>3-5x >=7 hoặc 3-5x>=-7

Xét tiếp...

\(VT=3\left(9x^2-12x+4\right)+\frac{8x}{1-x}=27x^2-36x+12+\frac{8x}{1-x}\)

\(=27x^2-36x+4+\frac{8}{1-x}=27x^2-18x-6+8\left(1-x\right)+\frac{8}{1-x}\)

\(=27x^2-18x+3+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(=3\left(3x-1\right)^2+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(\Rightarrow VT\ge2\sqrt{8^2}-9=7\)

Dấu " = " xảy ra khi \(x=\frac{1}{3}\)

Theo bài ra ta có:

|5x-3| lớn hơn hoặc bằng 7

=> 5x-3 lớn hơn hoặc bằng 7 hoặc 5x-3 lớn hơn hoặc bằng -7

=> x lớn hơn hoặc bằng 2 hoặc x lớn hơn hoặc bằng 4/15

PS mình ko ghi đc dấu lớn hơn hoặc bằng

Ta có: \(\left|5x-3\right|\ge7\)

\(\Rightarrow\orbr{\begin{cases}5x-3\ge7\\5x-3\ge-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x\ge10\\5x\ge-4\end{cases}\Rightarrow\orbr{\begin{cases}x\ge2\\x\ge-\frac{4}{5}\end{cases}}}\)

_Học tốt_