\(\left(2x+1\right)\left(4x^2-2x+1\right)-8x\left(x^2+2\right)=17\\ \Leftrightarrow\left(8x^3-4x^2+2x+4x^2-2x+1\right)-\left(8x^3+16x\right)=17\\ \Leftrightarrow\left(8x^3+1\right)-\left(8x^3+16x\right)=17\\ \Leftrightarrow16x+1=17\\ \Leftrightarrow x=1\)

\(A=\dfrac{2x^2-8x+17}{x^2-2x+1}\left(x\ne1\right)\)

\(\Leftrightarrow A\left(x^2-2x+1\right)=2x^2-8x+17\)

\(\Leftrightarrow Ax^2-2Ax+A=2x^2-8x+17\)

\(\Leftrightarrow x^2\left(A-2\right)-2x\left(A-4\right)+A-17=0\left(1\right)\)

\(A-2=0\Leftrightarrow A=2\Leftrightarrow x=3,75\left(tm\right)\left(2\right)\)

\(A-2\ne0\Leftrightarrow A\ne2\Rightarrow\Delta'\ge0\Leftrightarrow\left(A-4\right)^2-\left(A-17\right)\left(A-2\right)\ge0\Leftrightarrow A\ge\dfrac{18}{11}\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\left(tm\right)\left(3\right)\)

\(\left(2\right)và\left(3\right)\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\)

a đề sai hay sao mà vô nghiệm ?

b)Áp dụng BĐT Cauchy-Schwarz ta có:

\(VP^2=\left(\sqrt{2x+1}+\sqrt{17-2x}\right)^2\)

\(\le\left(1+1\right)\left(2x+1+17-2x\right)=36\)

\(\Rightarrow VP^2\le36\Rightarrow VP\le6\)

Lại có: \(VT=x^4-8x^3+17x^2-8x+22\)

\(=\left(x-4\right)^4+8\left(x-4\right)^3+17\left(x-4\right)^2+6\ge6\)

Thấy: \(VT\le VP=6\)\(\Rightarrow VT=VP=6\)

\(\Rightarrow\left(x-4\right)^4+8\left(x-4\right)^3+17\left(x-4\right)^2+6=6\)

Suy ra x=4

ko hiểu chỗ nào ib nhé

lời giải của bạn trên có 1 xíu sai nhé

Là BĐT Bu-nhi-a Cốp-xki chứ ạ ?