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a)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)
\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)
Vậy pt có một nghiệm duy nhất là \(x=-1\)
b)
\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)
\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)
\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)
Lập bảng xét dấu ra nhé ~^o^~
ĐKXĐ: \(\frac{4-\sqrt{10}}{2}\le x\le\frac{4+\sqrt{10}}{2}\)
Đặt : \(\sqrt{3x^2-12x+21}=a;\sqrt{5x^2-20x+24}=b\left(a,b>0\right)\Rightarrow a^2-b^2=-2x^2+8x-3\)
Khi đó pt trở thành:
\(a+b=a^2-b^2\)
\(\Rightarrow a=b\)
Theo cách đặt: \(\sqrt{3x^2-12x+21}=\sqrt{5x^2-20x+24}\)
\(\Leftrightarrow2x^2-8x+3=0\)
Đến đây bạn tự giải nha
+ \(\sqrt{3x^2-12x+21}=\sqrt{3\left(x-2\right)^2+9}\ge3\)
\(\sqrt{5x^2-20x+24}=\sqrt{5\left(x-2\right)^2+4}\ge2\)
=> \(VT\ge5\) Dấu "=" \(\Leftrightarrow x=2\) (1)
+ VP \(=-2\left(x^2-4x+4\right)+5=-2\left(x-2\right)^2+5\le5\forall x\) (2)
Dấu "=" \(\Leftrightarrow x=2\)
+ Từ (1) và (2) suy ra
\(pt\Leftrightarrow VT=VP=5\) \(\Leftrightarrow x=2\)
\(\sqrt{3x^2-12x+21}=\sqrt{3x^2-12x+12+9}=\sqrt{3\left(x-2\right)^2+9}\ge\sqrt{9}=3\)
\(\sqrt{5x^2-20x+24}=\sqrt{5x^2-20x+20+4}=\sqrt{5\left(x-2\right)^2+4}\ge\sqrt{4}=2\)
\(-2x^2+8x-3=-2x+8x-8+5=-2\left(x-2\right)^2+5\le5\)
\(VP\ge3+2=5,VT\le5\)
Suy ra \(VP=VT=5\)
Suy ra nghiệm của phương trình đạt tại \(x-2=0\Leftrightarrow x=2\).
a/ ĐKXĐ: ....
\(\Leftrightarrow2x^2+2x+4+2x-4=5\sqrt{\left(x-2\right)\left(x^2+x+2\right)}\)
\(\Leftrightarrow2\left(x^2+x+2\right)+2\left(x-2\right)=5\sqrt{\left(x-2\right)\left(x^2+x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a\\\sqrt{x-2}=b\end{matrix}\right.\)
\(\Leftrightarrow2a^2+2b^2=5ab\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\sqrt{x-2}\\2\sqrt{x^2+x+2}=\sqrt{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\left(x-2\right)\\4\left(x^2+x+2\right)=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+10=0\\4x^2+3x+10=0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: ....
\(\Leftrightarrow2x^2-x+1=\sqrt{4x^4+4x^2+1-4x^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2+1\right)^2-\left(2x\right)^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
\(\Leftrightarrow\frac{3}{4}\left(2x^2-2x+1\right)+\frac{1}{4}\left(2x^2+2x+1\right)=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2-2x+1}=a\\\sqrt{2x^2+2x+1}=b\end{matrix}\right.\)
\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\\3\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)
1/
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y-2=0\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=-3\end{matrix}\right.\)
2/ \(P=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(3-5x\right)^2}\)
\(P=\left|5x-2\right|+\left|3-5x\right|\ge\left|5x-2+3-5x\right|=1\)
\(\Rightarrow P_{min}=1\) khi \(\frac{2}{5}\le x\le\frac{3}{5}\)
3/ ĐKXĐ: \(\left|x\right|\ge1\)
\(x^2-1-\sqrt{x^2-1}=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\\sqrt{x^2-1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2-1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\right.\)