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a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\) => bpt vô nghiệm
b/ ĐKXĐ: \(x>1\)
\(bpt\Leftrightarrow x-2< 2\Leftrightarrow x< 4\)
\(\Rightarrow1< x< 4\)
c/ \(\frac{x+2}{3}-2x-2>0\)
\(\Leftrightarrow\frac{x+2-6x-6}{3}>0\Leftrightarrow x+2-6x-6>0\Leftrightarrow x< -\frac{4}{5}\)
d/ \(bpt\Leftrightarrow\frac{3x+5}{2}-\frac{x+2}{3}-x-1\le0\)
\(\Leftrightarrow\frac{9x+15-2x-4-6x-6}{6}\le0\)
\(\Leftrightarrow x\le-5\)
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:
\(5t< 2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2>0\Rightarrow\left[{}\begin{matrix}t>2\\t< \frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0\le x< \frac{3-2\sqrt{2}}{2}\\x>\frac{3+2\sqrt{2}}{2}\end{matrix}\right.\)
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1. ĐK x >1
pt \(\Leftrightarrow\frac{1}{\sqrt{x}-\sqrt{x-1}}\left(m\sqrt{x}+\frac{1}{\sqrt{x-1}}-16\sqrt[4]{\frac{x^3}{x-1}}\right)=1\)
\(\Leftrightarrow m\sqrt{x}+\frac{1}{\sqrt{x-1}}-16\sqrt[4]{\frac{x^3}{x-1}}=\sqrt{x}-\sqrt{x-1}\)
\(\Leftrightarrow m\sqrt{x\left(x-1\right)}+1-16\sqrt[4]{x^3\left(x-1\right)}=\sqrt{x\left(x-1\right)}-x+1\)
\(\Leftrightarrow\left(m-1\right)\sqrt{x\left(x-1\right)}-16\sqrt[4]{x^3\left(x-1\right)}+x=0\)
\(\Leftrightarrow\left(m-1\right)\sqrt{\frac{x-1}{x}}-16\sqrt[4]{\frac{x-1}{x}}+1=0\)
Đặt rồi đưa về phương trình bậc 2: \(\left(m-1\right)t^2-16t+1=0\)
2. ĐK:...
\(\sqrt{x-4-2\sqrt{x-4}+1}+\sqrt{x-4-2.\sqrt{x-4}.3+9}=m\)
\(\Leftrightarrow\left|\sqrt{x-4}-1\right|+\left|\sqrt{x-4}-3\right|=m\)Tìm m để pt có đúng 2 nghiệm. Tự làm nhé!
\(3.\) ĐK:...
Đặt: \(\left(x^2-3x-4\right)=a\)
\(\sqrt{x+7}=b\)
Ta có: \(ab-m\left(a-b\right)-m^2=0\Leftrightarrow m^2+m\left(a-b\right)-ab=0\)
\(\Delta=\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
pt có 2 nghiệm : \(\orbr{\begin{cases}m=\frac{b-a-\left(a+b\right)}{2}=-a\\m=\frac{b-a+\left(a+b\right)}{2}=b\end{cases}}\)
Khi đó: \(\orbr{\begin{cases}m=-\left(x^2-3x-4\right)\\m=\sqrt{x+7}\end{cases}}\)
pt <=> \(\left(m+x^2-3x-4\right)\left(m-\sqrt{x+7}\right)=0\)Tìm m để pt có nhiều nghiệm nhất .
a) ĐK : \(x\ge\frac{2}{3}\)\(\sqrt{3x-2}-\sqrt{x+7}=1\Leftrightarrow3x-2-2\sqrt{\left(3x-2\right)\left(x+7\right)}+x+7=1\)
\(\Leftrightarrow4x+5-1=2\sqrt{3x^2+19x-14}\Leftrightarrow2x+2=\sqrt{3x^2+19x-14}\)
\(\Leftrightarrow4x^2+8x+4=3x^2+19x-14\)
\(\Leftrightarrow x^2-11x+18=0\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
b) ĐK \(x\ge-\frac{1}{5}\)\(\sqrt{14x+7}-\sqrt{2x+3}=\sqrt{5x+1}\Leftrightarrow14x+7+2x+3-5x-1-2\sqrt{28x^2+42x+14x+21}=0\)
\(\Leftrightarrow11x+9=2\sqrt{28x^2+56x+21}\Leftrightarrow121x^2+81+198x=112x^2+224x+84\)
\(\Leftrightarrow9x^2-26x-3=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{1}{9}\left(loai\right)\end{matrix}\right.\)
c) \(\sqrt{x^2+2x+6}-\sqrt{x^2+x+2}=1\)
\(\Leftrightarrow x^2+2x+6=x^2+x+2+1+2\sqrt{x^2+x+2}\)
\(\Leftrightarrow x+3=2\sqrt{x^2+x+2}\)
\(\Leftrightarrow x^2+6x+9=4x^2+4x+8\)
\(\Leftrightarrow3x^2-2x-1=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\frac{1}{3}\left(tm\right)\end{matrix}\right.\)
a)\(ĐK:-3\le x\le6\)
\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)
\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)
b/ ĐKXĐ: \(x\ge7\)
\(\sqrt{3x-2}=1+\sqrt{x-7}\)
\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)
\(\Leftrightarrow x+2=\sqrt{x-7}\)
\(\Leftrightarrow x^2+4x+4=x-7\)
\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)
c/ ĐKXĐ: \(x\ge1;x\ne50\)
\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)
\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)
\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))
\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)