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![](https://rs.olm.vn/images/avt/0.png?1311)
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 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Coi như bước trên bạn đã làm đúng, giải pt vô tỉ thôi nhé:
TH1: \(x=y\)
\(\Rightarrow x^2+x+2=\sqrt{5x+5}+\sqrt{3x+2}\)
\(\Leftrightarrow x^2-x-1+\left(x+1-\sqrt{3x+2}\right)+\left(x+2-\sqrt{5x+5}\right)=0\)
\(\Leftrightarrow x^2-x-1+\dfrac{x^2-x-1}{x+1+\sqrt{3x+2}}+\dfrac{x^2-x-1}{x+2+\sqrt{5x+5}}=0\)
TH2: \(x=4y+3\)
Đây là trường hợp nghiệm ngoại lai, lẽ ra phải loại (khi bình phương lần 2 phương trình đầu, bạn quên điều kiện nên ko loại trường hợp này)
Dạ em cảm ơn thầy ạ, em ko nhìn ra cách chuyển thành x2 - x - 1 ạ @@
![](https://rs.olm.vn/images/avt/0.png?1311)
e/
ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow x^2+8x-2+6\sqrt{x\left(x+1\right)\left(x-2\right)}\le5x^2-4x-6\)
\(\Leftrightarrow3\sqrt{x\left(x+1\right)\left(x-2\right)}\le2x^2-6x-2\)
\(\Leftrightarrow3\sqrt{\left(x^2-2x\right)\left(x+1\right)}\le2x^2-6x-2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x}=a\ge0\\\sqrt{x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2-2b^2=2x^2-6x-2\)
BPT trở thành:
\(3ab\le2a^2-2b^2\Leftrightarrow2a^2-3ab-2b^2\ge0\)
\(\Leftrightarrow\left(2a+b\right)\left(a-2b\right)\ge0\)
\(\Leftrightarrow a\ge2b\Rightarrow\sqrt{x^2-2x}\ge2\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x\ge4x+4\)
\(\Leftrightarrow x^2-6x-4\ge0\)
\(\Rightarrow x\ge3+\sqrt{13}\)
d/
ĐKXĐ: \(x\ge-1\)
\(3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}+4x^2-5x+3\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow4a^2-b^2=4x^2-5x+3\)
BPT trở thành:
\(4a^2+3ab-b^2\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(4a-b\right)\ge0\)
\(\Leftrightarrow4a-b\ge0\Rightarrow4a\ge b\)
\(\Rightarrow4\sqrt{x^2+x+1}\ge\sqrt{x+1}\)
\(\Leftrightarrow16x^2+16x+4\ge x+1\)
\(\Leftrightarrow16x^2+15x+3\ge0\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le\frac{-15-\sqrt{33}}{32}\\x\ge\frac{-15+\sqrt{33}}{32}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ:\(x\ne-3\)
\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)
b, ĐKXĐ:\(x>2\)
\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x+3\right)\sqrt{2x^2+1}-\left(x+3\right)=x^2\)
=>\(\left(x+3\right)\cdot\left(\sqrt{2x^2+1}-1\right)=x^2\)
=>\(\left(x+3\right)\cdot\dfrac{2x^2+1-1}{\sqrt{2x^2+1}+1}-x^2=0\)
=>\(x^2\left(\dfrac{2\left(x+3\right)}{\sqrt{2x^2+1}+1}-1\right)=0\)
=>x^2=0 hoặc \(\dfrac{2\left(x+3\right)}{\sqrt{2x^2+1}+1}=1\)
=>\(\left[{}\begin{matrix}x=0\\\sqrt{2x^2+1}+1=2x+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2+1=\left(2x+5\right)^2;x>=-\dfrac{5}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\4x^2+20x+25-2x^2-1=0;x>=-\dfrac{5}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\\left\{{}\begin{matrix}2x^2+20x+24=0\\x>=-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5+\sqrt{13}\end{matrix}\right.\)
=>Phương trình này có 2 nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)