\(\Leftrightarrow\sqrt{12-7x}-\sqrt{x^2-x}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
\(\Rightarrow-\sqrt{3x^2-5x-1}-\sqrt{x^2-x}+\sqrt{x^2-3x+4}+\sqrt{12-7x}=0\)
=>\(x\approx-3,4579061804411\)
\(\Leftrightarrow\sqrt{12-7x}-\sqrt{x^2-x}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
\(\Rightarrow-\sqrt{3x^2-5x-1}-\sqrt{x^2-x}+\sqrt{x^2-3x+4}+\sqrt{12-7x}=0\)
=>\(x\approx-3,4579061804411\)
\(\sqrt{3x^2-7x+9x}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+13}\)
giải pt trên
\(\text{Giải pt :}\)
\(\sqrt{3x^2-7x+5}-\sqrt{x^2+2}=\sqrt{3x^2-5x+1}-\sqrt{x^2-x+4}\)
giải pt: \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
\(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
giải pt:
a)\(\sqrt{4x+1}\)-\(\sqrt{3x-2}\)=\(\frac{x+2}{5}\)
b)\(\sqrt{3x^2-7x+3}\)-\(\sqrt{x^2-2}\)=\(\sqrt{3x^2-5x-1}\)-\(\sqrt{x^2-3x+4}\)
Giải phương trình :
\(\sqrt{3x^2+7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
giai pt sau
\(\sqrt{3x-1}-\sqrt{x+2}.\sqrt{3x^2+7x+2}+4=4x-2\)
\(x^2-5x+3.\sqrt{2x-1}=2.\sqrt{14-2x}+5\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
\(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\) = \(\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
\(3x^3-17x^2-8x+9+\sqrt{3x-2}-\sqrt{7-x}\) = 0
GIẢI PHƯƠNG TRÌNH
giải pt
\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}\)= \(x+\sqrt{2x-\frac{5}{x}}\)
\(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\)\(\sqrt{x^2-3x+4}\)
\(\left(x+2\right)\sqrt{x+1}=2x+1\)
CỨU TÔI VỚI NGÀY MAI I HOK ÒI. PLEASE HELP HELP HELP ME