Giải pt x bình + 3 căn x bình + 3x = 10 - 3x
Giải pt :
1. trị tuyệt đối 5- 3x = 2
2. căn x bình - 6x + 9 = 3
3. căn ( x-1) bình = x + 3
Ghi chú : Cho em hỏi có ai rảnh kèm riêng giúp em môn này ko ạ
Giải pt:
1) Căn(x^2 - x + 2) + 1 = căn(10 - x^2 + x)
2) 4căn(x) - 2 căn(2 - x) + x - 4 căn( 2x - x^2) + 1 =0
3) x^2 + 3x - 1= (x+2) căn(x^2 + x - 1)
4) 3x^2 + 4x + 2 = 3(x+2) căn(x^2 - 1)
giải pt 6 căn(x+2) +3 căn(x-3)= căn(-x^2+x+6) +3x+1
giải pt: căn (x^2+4x+3)+căn (x^2+x)=căn (3x^2+4x+1)
\(\sqrt{x^2+4x+3}+\sqrt{x^2+x}=\sqrt{3x^2+4x+1}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}+\sqrt{x\left(x+1\right)}=\sqrt{\left(x+1\right)\left(3x+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}+\sqrt{x\left(x+1\right)}-\sqrt{\left(x+1\right)\left(3x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+3}+\sqrt{x}-\sqrt{3x+1}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+3}+\sqrt{x}=\sqrt{3x+1}\end{cases}}\)
Suy ra x=-1 pt còn lại bình lên là thấy vô nghiệm
Giải pt: 5[căn(3x-2)+căn(x+3)]=4x2-24x+35
Giải pt căn x + 7 + 6 căn x - 3x = 9 - căn 11- x
giải pt
(x-2)(căn 3x+1)-1=3x
(\(x\) - 2)(\(\sqrt{3x+1}\) ) - 1 = 3\(x\) Đk : 3\(x\) + 1 ≥ 0; \(x\) ≥ - \(\dfrac{1}{3}\)
(\(x\) - 2)(\(\sqrt{3x+1}\)) - (3\(x\) + 1) = 0
\(\sqrt{3x+1}\).(\(x\) - 2 - \(\sqrt{3x+1}\)) = 0
\(\left[{}\begin{matrix}\sqrt{3x+1}=0\\x-2-\sqrt{3x+1}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x-2=\sqrt{3x+1}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2-4x+4=3x+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2-7x+3=0\end{matrix}\right.\)
\(x^2\) - 7\(x\) + 3 = 0
△ = 49 -12 = 37
\(x_1\) = \(\dfrac{7+\sqrt{37}}{2}\)
\(x_{_{ }2}\) = \(\dfrac{-7-\sqrt{37}}{2}\) (loại)
Giải pt
Căn (x-3) ×(x^2 -3x+2) =0
giải pt: căn(x+3)+căn(3x+1)=2căn(x)+căn(2x+2)
Đk:\(x\ge0\)
\(\sqrt{x+3}+\sqrt{3x+1}=2\sqrt{x}+\sqrt{2x+2}\)
\(pt\Leftrightarrow\sqrt{x+3}-2+\sqrt{3x+1}-2=2\sqrt{x}-2+\sqrt{2x+2}-2\)
\(\Leftrightarrow\frac{x+3-4}{\sqrt{x+3}+2}+\frac{3x+1-4}{\sqrt{3x+1}-2}=\frac{4x-4}{2\sqrt{x}+2}+\frac{2x+2-4}{\sqrt{2x+2}+2}\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{3x-3}{\sqrt{3x+1}-2}=\frac{4x-4}{2\sqrt{x}+2}+\frac{2x-2}{\sqrt{2x+2}+2}\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{3\left(x-1\right)}{\sqrt{3x+1}-2}-\frac{4\left(x-1\right)}{2\sqrt{x}+2}-\frac{2\left(x-1\right)}{\sqrt{2x+2}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)