Giải pt :
\(3-x+\sqrt{x^2-3x+2}=0\)
K
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LD
1
TK
12 tháng 5 2021
\(3x^2+\sqrt{2}x-3+\sqrt{2}=0\)
Ta có \(a-b+c=3-\sqrt{2}-3+\sqrt{2}=0\)
Vậy phương trình có 2 nghiệm phân biệt
\(x_1=-1\)
\(x_2=-\dfrac{-3+\sqrt{2}}{3}=\dfrac{3-\sqrt{2}}{3}\)
HD
0
TK
2
ML
7 tháng 5 2018
\(pt\Leftrightarrow x^3-\sqrt{2}.x^2-2\sqrt{2}.x^2+4x-x+\sqrt{2}=0\)
\(\Leftrightarrow x^2\left(x-\sqrt{2}\right)-2\sqrt{2}x\left(x-\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x^2-2\sqrt{2}x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x^2-2\sqrt{2}x+2-3\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)[\left(x-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2]=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x-\sqrt{2}-\sqrt{3}\right)\left(x-\sqrt{2}+\sqrt{3}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=\sqrt{2}+\sqrt{3}\\x=\sqrt{2}-\sqrt{3}\end{cases}}\)
NT
2
23 tháng 10 2021
\(ĐK:-\dfrac{1}{3}\le x\le2\\ PT\Leftrightarrow\left(\sqrt{3x+1}-2\right)-x+1-\sqrt{2-x}\left(\sqrt{2-x}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}-\left(x-1\right)-\dfrac{\sqrt{2-x}\left(1-x\right)}{\sqrt{2-x}+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1=0\end{matrix}\right.\)
Với \(x\ge-\dfrac{1}{3}\) thì \(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{\sqrt{2-x}}{\sqrt{2-x}+1}-1>0\)
Vậy pt có nghiệm duy nhất \(x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
ĐKXĐ: \(-\dfrac{1}{3}\le x\le2\)
\(\sqrt{3x+1}=3-\sqrt{2-x}\) (do \(-\dfrac{1}{3}\le x\le2\Rightarrow3-\sqrt{2-x}\ge3-\sqrt{2+\dfrac{1}{3}}>0\))
\(\Leftrightarrow3x+1=9+2-x-6\sqrt{3-x}\)
\(\Leftrightarrow3\sqrt{2-x}=5-2x\)
\(\Leftrightarrow9\left(2-x\right)=\left(5-2x\right)^2\)
\(\Leftrightarrow4x^2-11x+7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\) (thỏa mãn)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
a.
\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:
\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)
\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)
\(\Leftrightarrow3a^2=b^2\)
\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
b.
\(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
Lặp lại cách làm câu a
MT
0
MT
2
DT
1
28 tháng 11 2021
a, ĐKXĐ: ...
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)
\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)
\(\Leftrightarrow4x^2-10x+3=0\)
.....
b, ĐKXĐ: ...
\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)
\(3-x+\sqrt{x^2-3x+2}=0\)
\(\Leftrightarrow\sqrt{x^2-3x+2}=x-3\)
\(ĐKXĐ:x^2-3x+2\ge0\)
\(\Leftrightarrow\left(x-2\right).\left(x-1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\ge1\end{matrix}\right.\)
\(pt\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\x^2-3x+2=\left(x-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x^2-3x+2=x^2-6x+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x=\dfrac{7}{3}\left(tm\right)\end{matrix}\right.\)
Vậy x = \(\dfrac{7}{3}\)
\(3-x+\sqrt{x^2-3x+2}=0\)
\(pt\Leftrightarrow\dfrac{7}{3}-x+\sqrt{x^2-3x+2}-\dfrac{2}{3}=0\)
\(\Leftrightarrow\dfrac{7}{3}-x+\dfrac{x^2-3x+2-\dfrac{4}{9}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)
\(\Leftrightarrow-\left(x-\dfrac{7}{3}\right)+\dfrac{\left(x-\dfrac{7}{3}\right)\left(x-\dfrac{2}{3}\right)}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{3}\right)\left(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}\right)=0\)
Dễ thấy: \(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}>0\forall\left[{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\)
\(\Rightarrow x-\dfrac{7}{3}=0\Rightarrow x=\dfrac{7}{3}\)