Giải pt:
\(\sqrt{2x+1}-\sqrt{3x}=x-1\)
Em cảm ơn ạ.
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V
1
AH
Akai Haruma
Giáo viên
18 tháng 6 2021
Nếu bạn thiếu số 2 bên cạnh $\sqrt{2x^2+5x+3}$ thì có thể tham khảo lời giải tại đây:
https://hoc24.vn/cau-hoi/tim-x-sao-cho-sqrt2x3sqrtx13x2sqrt2x25x3-16.235781793134
V
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2021
\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{2x-9}=\sqrt[3]{x-5}+\sqrt[3]{4x-3}\)
Đặt \(\sqrt[3]{3x+1}=a;\sqrt[3]{2x-9}=b;\sqrt[3]{x-5}=c;\sqrt[3]{4x-3}=d\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c+d\\\left(a+b\right)^3-3ab\left(a+b\right)=\left(c+d\right)^3-3cd\left(c+d\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+b=c+d=0\\\left[{}\begin{matrix}a+b=c+d\ne0\\ab=cd\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3+b^3=0\\a^3b^3=c^3d^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\\left(3x+1\right)\left(2x-9\right)=\left(4x-3\right)\left(x-5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\x^2-x-12=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
V
1
AT
20 tháng 6 2021
ĐKXĐ: \(x^3-1\ge0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)\ge0\)
mà \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)
\(2x^2+5x-1=7\sqrt{x^3-1}\Leftrightarrow2x^2+2x+2+3x-3=7\sqrt{x-1}\sqrt{x^2+x+1}\)
\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{x-1}\sqrt{x^2+x+1}\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{x^2+x+1}\end{matrix}\right.\left(a,b\ge0\right)\)
\(\Rightarrow\) pt trở thành \(2b^2+3a^2=7ab\Rightarrow2b^2-7ab+3a^2=0\)
\(\Rightarrow2b^2-6ab-ab+3a^2=0\Rightarrow2b\left(b-3a\right)-a\left(b-3a\right)=0\)
\(\Rightarrow\left(b-3a\right)\left(2b-a\right)=0\Rightarrow\left[{}\begin{matrix}b=3a\\2b=a\end{matrix}\right.\)
\(TH_1:b=3a\Rightarrow\sqrt{x^2+x+1}=3\sqrt{x-1}\)
\(\Rightarrow x^2+x+1=9\left(x-1\right)\Rightarrow x^2-8x+10=0\)
\(\Delta=\left(-8\right)^2-4.10=24\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{8-\sqrt{24}}{2}=4-\sqrt{6}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{8+\sqrt{24}}{2}=4+\sqrt{6}\end{matrix}\right.\)
\(TH_2:2b=a\Rightarrow2\sqrt{x^2+x+1}=\sqrt{x-1}\)
\(\Rightarrow4\left(x^2+x+1\right)=x-1\Rightarrow4x^2+3x+5=0\)
mà \(4x^2+3x+5=\left(2x\right)^2+2.2x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\dfrac{71}{16}=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}>0\)
\(\Rightarrow\) loại
Vậy pt có tập nghiệm \(S=\left\{4+\sqrt{6};4-\sqrt{6}\right\}\)
V
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2021
Đặt \(\sqrt{x^2+1}=t>0\)
\(\Rightarrow\left(4x-1\right)t=2t^2-2x\)
\(\Leftrightarrow2t^2-\left(4x-1\right)t-2x=0\)
\(\Delta=\left(4x-1\right)^2+16x=\left(4x+1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{4x-1-\left(4x+1\right)}{4}=-\dfrac{1}{2}\left(loại\right)\\t=\dfrac{4x-1+4x+1}{4}=2x\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+1}=2x\) (\(x\ge0\))
\(\Leftrightarrow x^2+1=4x^2\)
\(\Rightarrow x=\dfrac{\sqrt{3}}{3}\)
DT
0
19 tháng 6 2021
Đk:\(x\ge1;x\le-2\)
Đặt \(t=\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}\)
\(\Rightarrow t^2=\left(x-1\right)\left(x+2\right)\)
Pttt: \(t^2+4t=12\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
TH1: \(t=2\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-1\right)\left(x+2\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x^2+x-6=0\end{matrix}\right.\)\(\Rightarrow x=2\) (thỏa mãn)
TH2:\(t=-6\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=-6\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1< 0\\\left(x-1\right)\left(x+2\right)=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x^2+x-38=0\end{matrix}\right.\)\(\Rightarrow x=\dfrac{-1-3\sqrt{17}}{2}\) (thỏa mãn)
Vậy...
Đk \(x\ge0\)
Pt \(\Leftrightarrow\dfrac{2x+1-3x}{\sqrt{2x+1}+\sqrt{3x}}=x-1\)
\(\Leftrightarrow\dfrac{1-x}{\sqrt{2x+1}+\sqrt{3x}}+\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}+1\right)=0\)
\(\Leftrightarrow1-x=0\)( vì \(\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}+1>0\) với mọi \(x\ge0\))
\(\Leftrightarrow x=1\)
Vậy S={1}