Bài 1 ) \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
\(\Leftrightarrow\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+16}=-\left(x^2+2x+1\right)^2+6\)
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=-\left(x+1\right)^2+6\)
Ta có : \(\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{4}=2\)
\(\sqrt{5\left(x+1\right)^2+16}\ge\sqrt{16}=4\)
\(-\left(x+1\right)^2+6\le6\)
Ta có : \(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge6\)
Ta lại có : \(-\left(x+1\right)^2+6\le6\)
\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=6\)
Dấu "=" xảy ra khi và chỉ khi \(x+1=0\Leftrightarrow x=-1.\)
Nếu có lỗi sai chỉ với nhazzzzzzzzz
2 ) \(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)
\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)
\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)
Ta có : \(\left|2x+5\right|+\left|x-4\right|\ge\left|2x+5+4-x\right|=\left|x+9\right|\)
\(\Rightarrow\) \(\left|x+9\right|=\left|x+9\right|\)
Vậy phương trình vô số nghiệm
- Sửa bài giúp mk nhaz
Câu 3 : \(2x^2-8x-3\sqrt{x^2-4x-5}=12\) ĐK : \(x\ge5\)
\(\Leftrightarrow2\left(x^2-4x\right)-3\sqrt{x^2-4x-5}=12\)
Đặt \(x^2-4x=t\)
\(\Leftrightarrow2t-3\sqrt{t-5}=12\)
\(\Leftrightarrow-3\sqrt{t-5}=12-2t\)
\(\Leftrightarrow9\left(t-5\right)=144-48t+4t^2\)
\(\Leftrightarrow4t^2-57t+189=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=9\left(loại\right)\\t=5,25\end{matrix}\right.\)
Khi \(t=9\) thay vào , ta có :
\(x^2-4x-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{13}\\x=2-\sqrt{13}\left(loại\right)\end{matrix}\right.\)
Khi \(t=5,25\) thay vào , ta có :
\(x^2-4x=5,25\)
\(\Leftrightarrow x^2-4x-5,25=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{37}}{2}\\x=\dfrac{4-\sqrt{37}}{2}\end{matrix}\right.\left(loại\right)\)
Vậy x cuối cùng bằng \(2+\sqrt{13}\).
4 ) \(\dfrac{x^2-3x}{\sqrt{x-3}}=0\) ĐK : \(x\ne3\)
Để phân thức trên bằng 0 thì :
\(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy \(x=3.\)
Câu 5 ) \(\sqrt{x^2-x-2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+1\right)}-\sqrt{x-2}=0\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+1\right)}=\sqrt{x-2}\)
\(\Leftrightarrow\left(\sqrt{\left(x-2\right)\left(x+1\right)}\right)^2=\left(\sqrt{x-2}\right)^2\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=x-2\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)
Vậy \(x=2\)
b ) \(\sqrt{x^2-1}+1=x^2\) ĐK ; \(x\ge1\)
\(\Leftrightarrow\sqrt{x^2-1}+1-x^2=0\)
\(\Leftrightarrow\sqrt{x^2-1}-x^2+1=0\)
\(\Leftrightarrow\sqrt{x^2-1}-\left(x^2-1\right)=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(1-\sqrt{x^2-1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\1-\sqrt{x^2-1}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\\sqrt{x^2-1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\x^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)( loại \(-1;-\sqrt{2}\) )
Vậy ........
c ) \(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)}=-\sqrt{x^2+x-2}\left(vôlis\right)\)
Vậy pt vô nghiệm .
