\(\sqrt{x^2+2x+1}=x+2\)
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AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)
b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2$
$\Leftrightarrow |\sqrt{x-1}+1|=2$
$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$
$\Leftrightarrow x=2$ (tm)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)
d.
ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
HP
8 tháng 8 2021
a, ĐK: \(x\ge1\)
\(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{x^2-1}}+\sqrt{2x+2\sqrt{x^2-1}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}}+\sqrt{x-1+x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-\sqrt{x+1}\right)^2}+\sqrt{\left(\sqrt{x-1}+\sqrt{x+1}\right)^2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+1}-\sqrt{x-1}+\sqrt{x-1}+\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow2\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow x+1=2\)
\(\Leftrightarrow x=1\left(tm\right)\)
HP
8 tháng 8 2021
b, ĐK: \(x\ge-1+\sqrt{2},x\le-1-\sqrt{2}\)
Đặt \(\sqrt{x^2+2x-1}=t\left(t\ge0\right)\)
\(pt\Leftrightarrow2\left(1-x\right)t=t^2-4x\)
\(\Leftrightarrow t^2-4x+2xt-2t=0\)
\(\Leftrightarrow\left(t-2\right)\left(2x+t\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\\sqrt{x^2+2x-1}=-2x\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-1\pm\sqrt{6}\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 11 2019
a/ ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{x+1}+\sqrt{x}+2x+1+2\sqrt{x^2+x}-2=0\)
Đặt \(\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}\)
\(\Rightarrow a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{x}=1\)
Mà \(x\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x+1}\ge1\end{matrix}\right.\) \(\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1\)
Dấu "=" xảy ra khi và chỉ khi \(x=0\)
b/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0\)
Đặt \(\sqrt{x-2}-\sqrt{x+2}=a< 0\)
\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\) , pt trở thành:
\(a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=-2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-2}-\sqrt{x+2}=-2\)
\(\Leftrightarrow\sqrt{x-2}+2=\sqrt{x+2}\)
\(\Leftrightarrow x+2+4\sqrt{x-2}=x+2\)
\(\Leftrightarrow4\sqrt{x-2}=0\Rightarrow x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 11 2019
c/ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)
\(\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}\), ta được:
\(a^2-a-20=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{2x+3}-3+\sqrt{x+1}-2=0\)
\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{2}{\sqrt{2x+3}+3}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
\(\Rightarrow x=3\)
9 tháng 7 2023
a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)
\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)
=>PTVN
b:
\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)
=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)
\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)
PT có hai nghiệm là;
\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)
\(\sqrt{x^2+2x+1}=x+2\)
\(\Rightarrow\sqrt{\left(x+1\right)^2}=x+2\)
\(\Rightarrow x+1=x+2\)
\(\Rightarrow x+1+x+2=0\)
\(\Rightarrow2x+3=0\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)