GPT : \(x^2+6x+2=\left(2x+x\right)\sqrt{x^2+5}\)
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2021
ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2021
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
NC
1
3 tháng 9 2019
Trả lời :
Con a giai pt vế trái rồi nhân căn bình phương cả 2 vế
Con b cũng giải pt vế phải chuyển vế rồi bình phương cả 2 vế
Chắc vậy
k bt
CT
2
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2021
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2021
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
I
1
9 tháng 11 2021
\(ĐK:-5\le x\le3\)
Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:
\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm pt là ...
18 tháng 1
a: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-3\end{matrix}\right.\)
\(\sqrt{\left(x-2\right)\left(x+3\right)}=5\)
=>\(\sqrt{x^2+x-6}=5\)
=>\(x^2+x-6=25\)
=>\(x^2+x-31=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{-1+5\sqrt{5}}{2}\left(nhận\right)\\x=\dfrac{-1-5\sqrt{5}}{2}\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=x-5\)
=>\(\left|2x+3\right|=x-5\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3\right)^2=\left(x-5\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3-x+5\right)\left(2x+3+x-5\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(x+8\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left[{}\begin{matrix}x=-8\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\varnothing\)
c: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-6x+9}=x+7\)
=>\(\sqrt{\left(x-3\right)^2}=x+7\)
=>\(\left|x-3\right|=x+7\)
=>\(\left\{{}\begin{matrix}x+7>=0\\\left(x-3\right)^2=\left(x+7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\\left(x-3-x-7\right)\left(x-3+x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\-10\left(2x+4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-7\\x+2=0\end{matrix}\right.\)
=>x=-2
d: ĐKXĐ: x>=3/2
\(\sqrt{2x-3}=x-1\)
=>\(\left\{{}\begin{matrix}2x-3=\left(x-1\right)^2\\x>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=2x-3\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-4x+4=0\\x>=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>x=2
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
1/ \(\dfrac{5}{3}\le x\le\dfrac{7}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x-5}=a>0\\\sqrt{7-3x}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\17-6x=2b^2+3\\6x-7=2a^2+3\end{matrix}\right.\)
Mặt khác theo BĐT Bunhiacốpxki:
\(a+b=\sqrt{3x-5}+\sqrt{7-3x}\le\sqrt{\left(1+1\right)\left(3x-5+7-3x\right)}=2\)
\(\Rightarrow0< a+b\le2\)
Ta được hệ pt:
\(\left\{{}\begin{matrix}a^2+b^2=2\\\left(2b^2+3\right).a+\left(2a^2+3\right)b=2+8ab\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2ab^2+3a+2a^2b+3b-8ab-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-2\\2ab\left(a+b\right)+3\left(a+b\right)-8ab-2=0\end{matrix}\right.\)
\(\Rightarrow\left(\left(a+b\right)^2-2\right)\left(a+b\right)+3\left(a+b\right)-4\left(a+b\right)^2+6=0\)
\(\Leftrightarrow\left(a+b\right)^3-4\left(a+b\right)^2+\left(a+b\right)+6=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=-1< 0\left(l\right)\\a+b=2\\a+b=3>2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a+b=2\) , dấu "=" xảy ra khi và chỉ khi:
\(3x-5=7-3x\Rightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
2/ ĐKXĐ: \(x\ne\pm2\)
\(\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-\left(\dfrac{15}{x^2-4}+5\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-5.\left(\dfrac{x^2-1}{x^2-4}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{x^2-1}{x^2-4}\right)-4\left[\left(\dfrac{x^2-1}{x^2-4}\right)-\left(\dfrac{x+1}{x-2}\right)^2\right]=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)-4\left(\dfrac{x+1}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}-\dfrac{4\left(x+1\right)}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{4\left(x+1\right)}{x-2}\\\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=4\left(x^2+3x+2\right)\\x^2-3x+2=x^2+3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2+15x+6=0\\6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5+\sqrt{17}}{2}\\x=\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)
Em kiểm tra lại đề bài, chỗ \(2x+x\) bên vế phải