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ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow x^3-1+2x-1-\sqrt{3x-2}+x+1-\sqrt{x+3}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{4x^2-7x+3}{2x-1+\sqrt{3x-2}}+\frac{x^2+x-2}{x+1+\sqrt{x+3}}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{\left(x-1\right)\left(4x-3\right)}{2x-1+\sqrt{3x-2}}+\frac{\left(x-1\right)\left(x+2\right)}{x+1+\sqrt{x+3}}\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1+\frac{4x-3}{2x-1+\sqrt{3x-2}}+\frac{x+2}{x+1+\sqrt{x+3}}\right)\le0\)
\(\Leftrightarrow x-1\le0\) (ngoặc đằng sau luôn dương)
\(\Rightarrow x\le1\Rightarrow\frac{2}{3}\le x\le1\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=1\end{matrix}\right.\) \(\Rightarrow a+b=5\)
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: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\sqrt{x^2+4x+3m+1}=x+3\)
\(\Leftrightarrow x^2+4x+3m+1=\left(x+3\right)^2\)
\(\Leftrightarrow x^2+4x+3m+1=x^2+6x+9\)
\(\Leftrightarrow2x=3m-8\)
\(\Leftrightarrow x=\frac{3m-8}{2}\)
Với x=\(\frac{3m-8}{2}\Rightarrow\left(\frac{3m-8}{2}\right)^2+4\cdot\frac{3m-8}{2}+3m+1\ge0\)
\(\Leftrightarrow\frac{9m^2-48m+64}{4}+6m-16+3m+1\ge0\)
\(\Leftrightarrow9m^2-12m+4\ge0\)
\(\Leftrightarrow\left(3m-2\right)^2\ge0\)(luôn đúng)
Dấu "=" xảy ra <=> \(3m-2=0\Leftrightarrow m=\frac{2}{3}\)
\(\Rightarrow a=2;b=3\)
\(\Rightarrow4a^2+3b^2+7=4\cdot2^2+3\cdot3^2+7=50\)
\(A^2=9-x+2\sqrt{\left(2x+5\right)\left(4-3x\right)}\ge9-x\ge9-\frac{4}{3}=\frac{23}{3}\)
\(\Rightarrow A\ge\sqrt{\frac{23}{3}}\Rightarrow a+b=26\)