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\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (ĐK: \(x\ge\frac{-1}{2}\) )
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)
Vậy .....
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)(1)
Vì VT > 0 nên VP >0
\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow x\ge-\frac{1}{2}\)
Khi đó \(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)
Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))
ĐK: \(\orbr{\begin{cases}x>0\\x< -2\end{cases}}\)
\(pt\Leftrightarrow\left(x^2+2x\right)-\left(x+1\right)\sqrt{x^2+2x}+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(x+1\right)\sqrt{x^2+2x}+2\left(x+1\right)-4=0\)
Đặt \(\sqrt{x^2+2x}=A;x+1=B\left(A>0\right)\), phương trình trở thành:
\(A^2-AB+2B-4=0\)
\(\Leftrightarrow\left(A^2-4\right)+B\left(2-A\right)=0\)
\(\Leftrightarrow\left(A-2\right)\left(A+2-B\right)=0\Leftrightarrow\orbr{\begin{cases}A-2=0\\A-B+2=0\end{cases}}\)
Trở về phương trình đầu, ta có:
TH1: \(A=2\Rightarrow\sqrt{x^2+2x}=2\Rightarrow x^2+2x=4\Rightarrow\orbr{\begin{cases}x=\sqrt{5}-1\left(n\right)\\x=-\sqrt{5}-1\left(n\right)\end{cases}}\)
TH2: \(\sqrt{x^2+2x}-\left(x+1\right)=-2\Leftrightarrow\sqrt{x^2+2x}=x-1\)
ĐK: x > 1
\(pt\Rightarrow x^2+2x=x^2-2x+1\Rightarrow x=\frac{1}{4}\left(l\right)\)
KL: PT có nghiệm \(x=-\sqrt{5}-1\) và \(x=\sqrt{5}-1\)
ĐKXĐ: \(0\le x\le7;x\in R\)
Phương trình cho tương đương: \(2\sqrt{x}+\left(7-x\right)=\left(2+\sqrt{x}\right)\sqrt{7-x}\)
Đặt \(\sqrt{x}=a,\sqrt{7-x}=b\) với \(a,b\ge0\). Khi đó ta có phương trình:
\(2a+b^2=\left(2+a\right)b\Leftrightarrow b^2-2b+2a-ab=0\)
\(\Leftrightarrow\left(b-2\right)\left(b-a\right)=0\). Đến đây thì dễ rồi :)
\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
\(DKXD:x>0\)
\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(\frac{1}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{1}{2\left(x+1\right)}\right)=0\)
\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\left(\text{ }x-1\right)\left(x^2+x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy PT có 2 nghiệm \(x=1;x=3\)
tai sao cho xcan bac hai lai bang 2