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Câu 6:
\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3< 1\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2x^2-x+6x-3-2x^2+3x}{\left(2x-3\right)\left(2x-1\right)}\le0\\x^2+3< \left(1-3x\right)^2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}8x-3\le0\\x^2+3< 1-6x+9x^2\end{cases}\Leftrightarrow\hept{\begin{cases}8x-3\le0\\8x^2-6x-2< 0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{3}{8}\\\frac{-1}{4}x< x< \frac{1}{4}\end{cases}\Rightarrow}S\left(\frac{-1}{4};\frac{3}{8}\right)}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
4. đặt \(\sqrt[3]{x+24}=a\) và \(\sqrt{12-x}=b\)(b>=0)
==>ta có hệ pt
\(\int_{a^3+b^2=36}^{a+b=6}\)<=> \(\int_{a^3+\left(6-a\right)^2=36}^{b=6-a}\)<=> \(\int_{b=6-a}^{a^3+a^2-12a=0}\)<=> \(\int_{b=6-a}^{a\left(a^2+a-12\right)=0}\)<=>\(\int_{b=6-a}^{a\left(a+4\right)\left(a-3\right)=0}\)
đến đây bạn tự tìm a;b rufit hay vào tìm x là ok
3. \(\Leftrightarrow\sqrt[3]{2x^2}-\sqrt[3]{x+1}+\sqrt[3]{2x^2+1}-\sqrt[3]{x+2}=0\)
\(\Leftrightarrow\frac{2x^2-x-1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{2x^2-x-1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}=0\)
\(\Leftrightarrow2x^2-x-1=0\)
( do \(\frac{1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}>0\forall xTMĐK\))
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2=\frac{9}{8}\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{3}{4}\\x-\frac{1}{4}=-\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\) ( TM )