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\(1=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\ge\left(a+b\right)\left[\left(a+b\right)^2-\frac{3}{4}\left(a+b\right)^2\right]=\frac{\left(a+b\right)^3}{4}\)
\(\Rightarrow\left(a+b\right)^3\le4\Rightarrow a+b\le\sqrt[3]{4}\)
\(A=\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\le\sqrt{2\sqrt[3]{4}}\)
\("="\Leftrightarrow a=b=\frac{1}{\sqrt[3]{2}}\)
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)
5.
\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)
Câu 6 bạn coi lại đề
4.
ĐKXĐ: ...
Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)
\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)
\(\Rightarrow2x^2-ax-a^2=0\)
\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)
Bài 1:
Ta có: \(\left(2x^2+x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2+x-4-2x+1\right)\left(2x^2+x-4+2x-1\right)=0\)
\(\Leftrightarrow\left(2x^2-x-3\right)\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow\left(2x^2+2x-3x-3\right)\left(2x^2-2x+5x-5\right)=0\)
\(\Leftrightarrow\left[2x\left(x+1\right)-3\left(x+1\right)\right]\left[2x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\x-1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\x=1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=1\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{2};1;\frac{-5}{2}\right\}\)
\(x=\sqrt{\sqrt{2}-1}\Leftrightarrow x^2+1=\sqrt{2}\Leftrightarrow x^4+2x^2-1=0\)
\(Q=x^2\left(x^4+2x^2-1\right)+x\left(x^4+2x^2-1\right)+2019=2019\)
\(x=\sqrt{\sqrt{2}-1}\)\(\Leftrightarrow x^2=\sqrt{2}-1\)\(\Leftrightarrow x^2+1=\sqrt{2}\)\(\Leftrightarrow\left(x^2+1\right)^2=2\)
rồi chuyển vế
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)