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a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)
\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)
\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)
\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow x^2-2.x.2+2^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=2\)
\(x^2-2x+3=t\left(t\ge0\right)\)
\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)
\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)
\(\Leftrightarrow-5t^2+11t-2=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)
ĐK: \(x\ge\frac{1}{2}\)
Đặt \(t=\sqrt{2x-1}\Leftrightarrow x=\frac{t^2+1}{2}\)(ĐK: \(t\ge0\)) thay vao phương trình ta được:
\(\sqrt{\frac{t^2+1}{2}+4+3t}\)+\(\sqrt{\frac{t^2+1}{2}+12-5t}=7\sqrt{2}\)
\(\Leftrightarrow\sqrt{\frac{t^2+6t+9}{2}}+\sqrt{\frac{t^2-10t+25}{2}}=7\sqrt{2}\)
\(\Leftrightarrow\frac{\sqrt{\left(t+3\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(t-5\right)^2}}{\sqrt{2}}=7\sqrt{2}\)
\(\Leftrightarrow\frac{\left|t+3\right|+\left|t-5\right|}{\sqrt{2}}=7\sqrt{2}\)
\(\Leftrightarrow t+3+\left|t-5\right|=14\)(vì \(t\ge0\Rightarrow t+3>0\))
\(\Leftrightarrow t+\left|t-5\right|=11\)
Xét TH: \(t-5\ge0\Leftrightarrow t\ge5\) thì ta có:
\(t+t-5=11\)
\(\Leftrightarrow2t=16\)
\(\Leftrightarrow t=8\)(chọn)
Xét TH: \(t-5< 0\Leftrightarrow t< 5\) thì ta có:
\(t-t+5=11\)
\(\Leftrightarrow5=11\)(vô lí nên loại)
Lại có: \(t=8\)
\(\Leftrightarrow\sqrt{2x-1}=8\)
\(\Leftrightarrow2x-1=64\)
\(\Leftrightarrow2x=63\)
\(\Leftrightarrow x=\frac{63}{2}=31\frac{1}{2}\)
Vậy nghiệm của phương trình là x=31\(\frac{1}{2}\)
\(\Rightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{2x^3}{2}+\frac{x^2}{2}+\frac{2x}{2}+\frac{1}{2}\)
\(\Rightarrow\sqrt{x^2+x+\frac{1}{2}-\frac{1}{4}}=\sqrt{x^2+x+\frac{1}{4}}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x+\frac{1}{2}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)
\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)
vậy x=0 và x=-1/2
ĐKXĐ: \(x\ne\left\{0;\frac{1}{2}\right\}\)
\(\Leftrightarrow\frac{3+x}{2x-1}-\frac{2x}{x\left(2x-1\right)}-3-\frac{x-4}{x}=0\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(2x-1\right)}-\frac{2x}{x\left(2x-1\right)}-\frac{3x\left(2x-1\right)}{x\left(2x-1\right)}-\frac{\left(x-4\right)\left(2x-1\right)}{x\left(2x-1\right)}=0\)
\(\Leftrightarrow x^2+3x-2x-6x^2+3x-2x^2+9x-4=0\)
\(\Leftrightarrow-7x^2+13x-4=0\)
\(\Rightarrow x=\frac{13\pm\sqrt{57}}{14}\)
Kết quả xấu quá, chắc bạn ghi ko đúng đề