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\(Pt\Leftrightarrow\left(x^3+4x^2+3x\right)+3\left(x+2-\sqrt[3]{2x^2+9x+8}\right)=0.\)
\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{\left(x+2\right)^3-2x^2-9x-8}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{2x^2+9x+8}+\sqrt[3]{\left(2x^2+9x+8\right)^2}}=0\)
\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{x^3+4x^2+3x}{MS}=0\Leftrightarrow\left(x^3+4x^2+3x\right)\left(1+\frac{3}{MS}\right)=0\)
Dễ thấy MS >0 \(\Rightarrow PT\Leftrightarrow x^3+4x^2+x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)
\(\Rightarrow Pt\Leftrightarrow x^3+4x^2+3x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)\(\Rightarrow PT\Leftrightarrow x^3+4x^2+3x=0\)<=>\(x\in\left\{-3;-1;0\right\}\)
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a) \(\sqrt{4x^2}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\4x^2=\left(x+1\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2-2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left(3x+1\right)\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=-\frac{1}{3}\\x=1\end{matrix}\right.\left(TM\right)\end{matrix}\right.\)
b) \(\sqrt{16x^2}=8\Leftrightarrow16x^2=64\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\) ( TM )
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\(\sqrt{27-8\sqrt{11}}-\sqrt{11}=\sqrt{\left(4-\sqrt{11}\right)^2}-\sqrt{11}=\left|4-\sqrt{11}\right|-\sqrt{11}=4-2\sqrt{11}\)
\(\sqrt{48-16\sqrt{8}}-\sqrt{8}=\sqrt{\left(4\sqrt{2}-4\right)^2}-2\sqrt{2}=\left|4\sqrt{2}-4\right|-2\sqrt{2}=2\sqrt{2}-4\)
\(\sqrt{27-8\sqrt{11}}-\sqrt{11}\\ =\sqrt{4^2-2.4.\sqrt{11}+\left(\sqrt{11}\right)^2}-\sqrt{11}\\ =\sqrt{\left(4-\sqrt{11}\right)^2}-\sqrt{11}\\ =\left|4-\sqrt{11}\right|-\sqrt{11}\\ =4-\sqrt{11}-\sqrt{11}=4-\left(-2\right)=6\)
\(\sqrt{48-16\sqrt{8}}-\sqrt{8}\\ =\sqrt{\left(4\sqrt{2}\right)^2-2.4\sqrt{2}.4+4^2}-\sqrt{8}\\ =\sqrt{\left(4\sqrt{2}-4\right)^2}-\sqrt{8}\\ =\left|4\sqrt{2}-4\right|-\sqrt{8}\\ =4\sqrt{2}-4-\sqrt{8}=4\sqrt{2}-4-3\sqrt{2}\\ =\sqrt{2}-4\)
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Lời giải:
Đặt $\sqrt{y}=b(b\geq 0)\Rightarrow y=b^2$
$M=2x^2+5b^2-4xb-4x-8b+2036$
$=2(x^2+b^2-2xb)+3b^2-4x-8b+2036$
$=2(x-b)^2-4(x-b)+3b^2-12b+2036$
$=2(x-b)^2-4(x-b)+2+3(b^2-4b+4)+2022$
$=2[(x-b)^2-2(x-b)+1]+3(b-2)^2+2022$
$=2(x-b-1)^2+3(b-2)^2+2022\geq 2022$
Vậy $M_{\min}=2022$
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Câu 1/ Ta có:
\(\left\{{}\begin{matrix}\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge1\\\sqrt{x^2-4x+8}=\sqrt{\left(x-2\right)^2+4}\ge2\\\sqrt{x^2-4x+9}=\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow VT\ge1+2+\sqrt{5}=VP\)
Dấu = xảy ra khi x = 2
PS: Câu còn lại thì chỉ cần phân tích cái trong căn thành số chính phương là xong.
Câu 2/ Sửa đề
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
Điều kiện: \(x\ge1\)
\(\Leftrightarrow\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)
Tới đây thì đơn giản rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
b, bạn kiểm tra lại đề nhé
c, \(\frac{x\sqrt{x}-8+2x-4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{x-4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-4\right)}{x-4}=\sqrt{x}+2\)
Mình nghĩ là không
\(-4x+8\sqrt{x}+8=-4\left(\sqrt{x}-1+\sqrt{3}\right)\left(\sqrt{x}-1-\sqrt{3}\right)\)