Nếu chỉ cần biện luận số nghiệm thì:
Đặt \(x^2=t\ge0\) \(\Rightarrow\left(2-\sqrt{5}\right)t^2+5t+7\left(1+\sqrt{2}\right)=0\) (1)
Ta có \(ac=\left(2-\sqrt{5}\right).7\left(1+\sqrt{2}\right)< 0\) nên (1) có 2 nghiệm trái dấu hay có đúng 1 nghiệm dương
\(\Rightarrow\) Pt đã cho có 2 nghiệm pb
giai pt
a) \(\sqrt{1+\sqrt{1-x^2}.}[\sqrt{\left(1-x\right)^3}-\sqrt{\left(1+x\right)^3}]=2+\sqrt{1-x^2}\)
b) \(\sqrt{1-x}-2x\sqrt{1-x^2}-2x^2+1=0\)
c) \(64x^6-112x^4+56x^2-7=2\sqrt{1-x^2}\)
\(\left\{{}\begin{matrix}y^3-4y^2+4y=\sqrt{x+1}\left(y^2-5y+4+\sqrt{x+1}\right)\\2\sqrt{x^2-3x+3}+6x-7=y^2\left(x-1\right)^2+\left(y^2-1\right)\sqrt{3x-2}\end{matrix}\right.\)
1) \(\left|x^2-4x-5\right|=x-1\)
2) \(\sqrt{2x^2+2x+9}=x-3\)
3) \(\sqrt{x+1}+1=4x^2+\sqrt{3x}\)
4) \(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-3\)
Giải các hệ
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{2x+y+2}=7\\3x+2y=23\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x^2+1\right)+y\left(x+y\right)=7y\\\left(x^2+1\right)\left(x+y-2\right)=-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\left(x+y+1\right)=3\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{matrix}\right.\)
1) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x^2+y^2-3xy+3x-2y+1=0\\4x^2-y^2+x+4=\sqrt{2x+y}+\sqrt{x+4y}\end{matrix}\right.\)
giải pt \(\left(4x+2\right)\sqrt{x^2+2x+5}=x^2+2x+5\sqrt{4x+5}\)
\(\begin{cases}2\sqrt{5-x-y}+2x=3\sqrt{\left(x+1\right)\left(2-y+y\right)}\\\left(x-y\right)^2+x+y=2\end{cases}\left(x,y\in R\right)\)
\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}=4\)
giải phương trình bằng cách đặt ẩn phụ
\(\left\{{}\begin{matrix}x\left(x+21\right)+y\left(x-33\right)=2\left(y^2+50\right)\\\sqrt{x+2}+2\sqrt{y+11}=\sqrt{\left(4y-x+14\right)^3}\end{matrix}\right.\)
