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 ?

Đk:\(-\frac{1}{2}\le x\le1\)

\(pt\Leftrightarrow3\sqrt[3]{2x+1}-3+\sqrt{1-x}-1=0\)

\(\Leftrightarrow\frac{27\left(2x+1\right)-27}{\left(3\sqrt[3]{2x+1}\right)^2+3\cdot3\sqrt[3]{2x+1}+3^2}+\frac{1-x-1}{\sqrt{1-x}+1}=0\)

\(\Leftrightarrow\frac{54x}{\left(3\sqrt[3]{2x+1}\right)^2+3\cdot3\sqrt[3]{2x+1}+3^2}+\frac{-x}{\sqrt{1-x}+1}=0\)

\(\Leftrightarrow x\left(\frac{54}{\left(3\sqrt[3]{2x+1}\right)^2+3\cdot3\sqrt[3]{2x+1}+3^2}+\frac{-1}{\sqrt{1-x}+1}\right)=0\)

suy ra x=0

thay điểm A vào đồ thị hàm số thì ta có:

3=-2+b

=>b=5

\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)

\(pt\Leftrightarrow4x^2+12+\sqrt{x-1}=4x\sqrt{5x-1}+4\sqrt{9-5x}\)

\(\Leftrightarrow4x^2-4+\sqrt{x-1}=4x\sqrt{5x-1}-8+4\sqrt{9-5x}-8\)

\(\Leftrightarrow4\left(x^2-1\right)+\sqrt{x-1}=\frac{16x^2\left(5x-1\right)-64}{4x\sqrt{5x-1}+8}+\frac{16\left(9-5x\right)-64}{4\sqrt{9-5x}+8}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)+\frac{x-1}{\sqrt{x-1}}=\frac{80x^3-16x^2-64}{4x\sqrt{5x-1}+8}+\frac{80-80x}{4\sqrt{9-5x}+8}\)

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)+\frac{x-1}{\sqrt{x-1}}-\frac{16\left(x-1\right)\left(5x^2+4x+4\right)}{4x\sqrt{5x-1}+8}+\frac{80\left(x-1\right)}{4\sqrt{9-5x}+8}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4\left(x+1\right)+\frac{1}{\sqrt{x-1}}-\frac{16\left(5x^2+4x+4\right)}{4x\sqrt{5x-1}+8}+\frac{80}{4\sqrt{9-5x}+8}\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

\(\sqrt{x+1}-\sqrt{x-2}=1\)

\(pt\Leftrightarrow\sqrt{x+1}-2-\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\frac{x+1-4}{\sqrt{x+1}+2}-\frac{x-2-1}{\sqrt{x-2}+1}=0\)

\(\Leftrightarrow\frac{x-3}{\sqrt{x+1}+2}-\frac{x-3}{\sqrt{x-2}+1}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x+1}+2}-\frac{1}{\sqrt{x-2}+1}\right)=0\)

Suy ra x=3

Đk:\(-2\le x\le3\)

\(pt\Leftrightarrow\sqrt{x+2}-\left(\frac{1}{3}x+\frac{4}{3}\right)+\sqrt{3-x}-\left(-\frac{1}{3}x+\frac{5}{3}\right)=0\)

\(\Leftrightarrow\frac{x+2-\left(\frac{1}{3}x+\frac{4}{3}\right)^2}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{3-x-\left(-\frac{1}{3}x+\frac{5}{3}\right)^2}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow\frac{x+2-\frac{x^2+8x+16}{9}}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{3-x-\frac{x^2-10x+25}{9}}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow\frac{-\frac{x^2-x-2}{9}}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{-\frac{x^2-x-2}{9}}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}=0\)

\(\Leftrightarrow-\frac{x^2-x-2}{9}\left(\frac{1}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{1}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}\right)=0\)

\(\Leftrightarrow-\frac{\left(x+1\right)\left(x-2\right)}{9}\left(\frac{1}{\sqrt{x+2}+\frac{1}{3}x+\frac{4}{3}}+\frac{1}{\sqrt{3-x}+\left(-\frac{1}{3}x+\frac{5}{3}\right)}\right)=0\)

Suy ra x=-1;x=2