giải
\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)
Đặt a = \(\sqrt{x^2+2x-1}\left(a\ge0\right)\) , ta đc pt: 2(1 - x).a = a2 - 4x => a2 - 2(1 - x)a - 4x = 0
Ta có: \(\Delta'=\left[-\left(1-x\right)\right]^2+4x=1-2x+x^2+4x=x^2+2x+1=\left(x+1\right)^2\)\(\Rightarrow\sqrt{\Delta'}=x+1\)
\(\Rightarrow\left[\begin{array}{nghiempt}a=\frac{1-x+x+1}{1}=2\\a=\frac{1-x-x-1}{1}=-2x\left(vn\right)\end{array}\right.\)
+) Với a = 2 \(\Rightarrow\sqrt{x^2+2x-1}=2\Rightarrow x^2+2x-1=4\Rightarrow x^2+2x-5=0\Rightarrow\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)
Vậy pt có 2 nghiệm \(\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)
ĐK:...
\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)
\(\Leftrightarrow2\left(1-x\right)\sqrt{\left(1+x\right)^2-2}=\left(1-x\right)^2-2\)
Đặt \(\begin{cases}a=1+x\\b=1-x\end{cases}\),ta có hệ:
\(\begin{cases}2b\sqrt{a^2-2}=b^2-2\\a+b=2\end{cases}\)
\(\Leftrightarrow\begin{cases}4a^2b^2-8b^2=b^4-4b^2+4\\a+b=2\end{cases}\)
\(\Leftrightarrow\begin{cases}4a^2b^2=b^4+4b^2+4\\a+b=2\end{cases}\)
\(\Leftrightarrow\begin{cases}2ab=b^2+2\\b=2-a\end{cases}\)hay\(\begin{cases}2ab=-b^2-2\\b=2-a\end{cases}\)
\(\Leftrightarrow2a\left(2-a\right)=\left(2-a\right)^2+2\)hay\(2a\left(2-a\right)=-\left(2-a\right)^2-2\)
\(\Leftrightarrow3a^2-8a+6=0\)hay a2=6
\(\Rightarrow\left[\begin{array}{nghiempt}a=x+1=\sqrt{6}\\a=x+1=-\sqrt{6}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)
ê ku vĩ căn đen ta =x+1 thế GTTĐ đâu quên cơ bản ak
Điều kiện x,y thuộc R hình như k liên quan thì phải
nghiệm đây chúng mày tham khảo xong khỏi cãi nhau
\(x=-\sqrt{6}-1;x=\sqrt{6}-1\)
ĐKXĐ:\(\left[\begin{array}{nghiempt}x\ge-1+\sqrt{2}\\x\le-1-\sqrt{2}\end{array}\right.\)
\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)
\(\Leftrightarrow x^2+2x-5+2\left(1-x\right).2=2\left(1-x\right)\sqrt{x^2+2x-1}\)
\(\Leftrightarrow x^2+2x-5=2\left(1-x\right)\left(\sqrt{x^2+2x-1}-2\right)\)
\(\Leftrightarrow x^2+2x-5=2\left(1-x\right)\frac{x^2+2x-5}{\sqrt{x^2+2x-1}+2}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+2x-5=0\\\sqrt{x^2+2x-1}=-2x\end{array}\right.\)
\(\Leftrightarrow x^2+2x-5=0\)
\(\Leftrightarrow x=-1\pm\sqrt{6}\)
Vậy pt có 2 nghiệm \(x=-1\pm\sqrt{6}\)
ĐKXĐ:....
\(\Leftrightarrow4\left(1-x\right)^2\left(x^2+2x-1\right)=\left(x^2-2x-1\right)^2\)
\(\Leftrightarrow3x^4+4x^3-18x^2+12x-5=0\)
\(\Leftrightarrow\left(x^2+2x-5\right)\left(3x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+2x-5=0\Rightarrow x=-1\pm\sqrt{6}\\3x^2-2x+1=0\left(false\right)\end{array}\right.\)
