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a) Đặt \(a=\sqrt[3]{1+\sqrt{x}};b=\sqrt[3]{1-\sqrt{x}}\)
\(\Rightarrow a^3+b^3=2\) kết hợp với đề bài
\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a^2-ab+b^2\right)=2\\a+b=2\end{matrix}\right.\)
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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:
\(4a^2+3ab-b^2\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(4a-b\right)\ge0\)
\(\Leftrightarrow4a-b\ge0\Rightarrow4a\ge b\)
\(\Rightarrow4\sqrt{x^2+x+1}\ge\sqrt{x+1}\)
\(\Leftrightarrow16x^2+16x+4\ge x+1\)
\(\Leftrightarrow16x^2+15x+3\ge0\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le\frac{-15-\sqrt{33}}{32}\\x\ge\frac{-15+\sqrt{33}}{32}\end{matrix}\right.\)
1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
\(\sqrt[3]{5x+7}+\sqrt[3]{5x-13}=1\)
☘ Đặt \(\sqrt[3]{5x+7}=a\text{ và }\sqrt[3]{5x-13}=b\), ta được hệ phương trình:
\(\left\{{}\begin{matrix}a+b=1\\a^3-b^3=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\\\left(1-b\right)^3-b^3=20\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(1-3b+3b^2-b^3\right)-b^3=20\)
\(\Leftrightarrow2b^3-3b^2+3b+19=0\)
⚠ Làm tiếp nhé.
Hương Thảo
\(\left\{{}\begin{matrix}\sqrt[3]{5x+7}=a\\\sqrt[3]{5x-13}=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+7=a^3\\5x-13=b^3\end{matrix}\right.\)
Trừ vế theo vế, ta được:
\(20=a^3-b^3\)
Okay?