1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
1) => 9(x-1)=\(21^2\)
=> 9x-9=441
=> 9x=450
=> x=50
2)=>\(\sqrt{1-x}\) + \(\sqrt{4\left(1-x\right)}\)-\(\dfrac{1}{3}\sqrt{16\left(1-x\right)}\)+5=0
=>\(\sqrt{1-x}\)\(\left(1+2-\dfrac{1}{3}.4\right)\)+5=0
=>\(\dfrac{5}{3}\sqrt{1-x}\) +5=0
=>\(\sqrt{1-x}\)=-3
Phuong trinh vo nghiem
1)
\(\sqrt{9\left(x-1\right)}=21\left(x\ge1\right)\\ < =>9\left(x-1\right)=441\\ < =>x-1=49\\ < =>x=50\left(tm\right)\)
2)
\(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\left(x\le1\right)\)
\(< =>\sqrt{1-x}+\sqrt{4\left(1-x\right)}-\dfrac{1}{3}\sqrt{16\left(1-x\right)}+5=0\\ < =>\sqrt{1-x}+2\sqrt{1-x}-\dfrac{1}{3}\cdot4\sqrt{1-x}=-5\\ < =>\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}=-5\\ < =>\dfrac{5}{3}\sqrt{1-x}=-5\\ < =>\sqrt{1-x}=-3\left(vl\right)\)
3)
\(\sqrt{2x}-\sqrt{50}=0\left(x\ge0\right)\\ < =>\sqrt{2x}=\sqrt{50}\\ < =>2x=50\\ < =>x=25\left(tm\right)\)
4)
\(\sqrt{4x^2+4x+1}=6\\ < =>\sqrt{\left(2x+1\right)^2}=6\\ < =>\left|2x+1\right|=6\\ < =>\left[{}\begin{matrix}2x+1=6\left(2x+1\ge0< =>x\ge-\dfrac{1}{2}\right)\\-2x-1=6\left(2x+1< 0< =>x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
với `x>=-1/2`
`2x+1=6`
`<=>2x=5`
`<=>x=5/2(tm)`
với `x<-1/2`
`-2x-1=6`
`<=>-2x=7`
`<=>x=-7/2(tm)`
5)
\(\sqrt{\left(x-3\right)^2}=3-x\\ < =>\left|x-3\right|=3-x\\ < =>\left[{}\begin{matrix}x-3=3-x\left(x-3\ge0< =>x\ge3\right)\\3-x=3-x\left(x-3< 0< =>x< 3\right)\end{matrix}\right.\)
với `x>=3`
`x-3=3-x`
`<=>2x=6`
`<=>x=3`
với `x<3`
`3-x=3-x`
`<=>0x=0` (luôn đúng)
4) \(\sqrt[]{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt[]{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
2) \(\sqrt[]{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\left(đk:x\le3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\x-3=-3+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\0x=0\left(luôn.đúng\right)\end{matrix}\right.\)
\(\)\(\Leftrightarrow\forall x\inℝ\) (so với \(x\le3\))
\(\Leftrightarrow x\le3\)
