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ĐK : \(x\ne-\frac{1}{2}\);\(x\ge0\)
\(\frac{\left(2x^2+8x+1\right)^2}{\left(2x+1\right)^2}=25x\)
\(25\left(4x^2+4x+1\right)=\left(4x^4+64x^2+1+32x^3+4x^2+16x\right)\)
\(4x^4+32x^3-32x^2-84x-24=0\)
giải tiếp đc nghiệm
1) ĐK: \(x\ge-1\)
\(\sqrt{9x^2+9x+4}>9x+3-\sqrt{x+1}\)
<=> \(\sqrt{9x^2+9x+4}+\sqrt{x+1}>9x+3\)(1)
TH1: 9x + 3 \(\le\)0 <=> x\(\le-\frac{1}{3}\)
(1) luôn đúng
Th2: x\(>-\frac{1}{3}\)
<=> \(\left(\frac{1}{2}x+1-\sqrt{x+1}\right)+\left(\frac{17}{2}x+2-\sqrt{9x^2+9x+4}\right)< 0\)
<=> \(\frac{\frac{1}{4}x^2}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{\frac{253}{4}x^2}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}< 0\)
<=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)< 0\)vô nghiệm
Vì với x \(>-\frac{1}{3}\):
ta có: \(\frac{1}{2}x+1+\sqrt{x+1}>0\)
\(\frac{17}{2}x+2+\sqrt{9x^2+9x+4}=\frac{17}{2}x+2+\sqrt{3\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}>\frac{17}{2}x+2+1>0\)
=> \(\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)>0\)với x \(>-\frac{1}{3}\) và \(x^2\ge0\)với mọi x
=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)\ge0\)với x\(>-\frac{1}{3}\)
Vậy \(x< -\frac{1}{3}\)
Xin lỗi bạn kết luận bài 1 là:
\(-1\le x\le-\frac{1}{3}\)
Bài 2) \(2+\sqrt{x+2}-x\sqrt{x+2}=x\left(\sqrt{x+2}-x\right)\)(2)
ĐK: \(x\ge-2\)
(2) <=> \(2+\sqrt{x+2}+x^2-2x\sqrt{x+2}=0\)
<=> \(8+4\sqrt{x+2}+4x^2-8x\sqrt{x+2}=0\)
<=> \(\left(2x-1\right)^2-4\left(2x-1\right)\sqrt{x+2}+4\left(x+2\right)-1=0\)
<=> \(\left(2x-1-2\sqrt{x+2}\right)^2-1=0\)
<=> \(\left(x-1-\sqrt{x+2}\right)\left(x-\sqrt{x+2}\right)=0\)
<=> \(\orbr{\begin{cases}x-1=\sqrt{x+2}\left(3\right)\\x=\sqrt{x+2}\left(4\right)\end{cases}}\)
(3) <=> \(\hept{\begin{cases}x\ge1\\x^2-3x-1=0\end{cases}}\Leftrightarrow x=\frac{3+\sqrt{13}}{2}\left(tm\right)\)
(4) <=> \(\hept{\begin{cases}x\ge0\\x^2-x-2=0\end{cases}\Leftrightarrow}x=2\left(tm\right)\)
Kết luận:...
a, ĐK: \(6x^2-12x+7\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)
Vậy...
a/ ĐKXĐ: ...
\(\Leftrightarrow4x^2-4x+1-\left(2x-\sqrt{4x-1}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\frac{\left(2x-1\right)^2}{2x+\sqrt{4x-1}}=0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(1-\frac{1}{2x+\sqrt{4x-1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\2x+\sqrt{4x-1}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x-1}=1-2x\) (\(x\le\frac{1}{2}\))
\(\Leftrightarrow4x-1=\left(1-2x\right)^2\)
\(\Leftrightarrow4x-1=4x^2-4x+1\)
\(\Leftrightarrow2x^2-4x+1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{2}}{2}\left(l\right)\\x=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\)
b/
Đặt \(3x^2-2x+2=a>0\) ta được:
\(\sqrt{a+7}+\sqrt{a}=7\)
\(\Leftrightarrow2a+7+2\sqrt{a^2+7a}=49\)
\(\Leftrightarrow\sqrt{a^2+7a}=21-a\) (\(a\le21\))
\(\Leftrightarrow a^2+7a=\left(21-a\right)^2\)
\(\Leftrightarrow a^2+7a=a^2-42a+441\)
\(\Rightarrow a=9\Rightarrow3x^2-2x+2=9\)
\(\Leftrightarrow3x^2-2x-7=0\Rightarrow x=\frac{1\pm\sqrt{22}}{3}\)
a/ ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow3\sqrt{x+8}\ge3\left(\sqrt{x+3}+\sqrt{x}\right)\)
\(\Leftrightarrow\sqrt{x+8}\ge\sqrt{x+3}+\sqrt{x}\)
\(\Leftrightarrow x+8\ge2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x\ge2\sqrt{x^2+3x}\)
- Với \(x>5\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP\ge0\end{matrix}\right.\) BPT vô nghiệm
- Với \(x\le5\) hai vế ko âm, bình phương:
\(x^2-10x+25\ge4x^2+12x\)
\(\Leftrightarrow3x^2+22x-25\le0\Rightarrow-\frac{25}{3}\le x\le1\)
Vậy nghiệm của BPT đã cho là \(0\le x\le1\)
b/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)< 2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\Rightarrow x+\frac{1}{4x}=t^2-1\)
BPT trở thành:
\(5t< 2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2>0\Rightarrow\left[{}\begin{matrix}t>2\\t< \frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0\le x< \frac{3-2\sqrt{2}}{2}\\x>\frac{3+2\sqrt{2}}{2}\end{matrix}\right.\)