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\(A=0.5\cdot4\sqrt{3-x}-\sqrt{3-x}-2\sqrt{3}+1=\sqrt{3-x}-2\sqrt{3}+1\) (xác định khi x=<3)
a)thay \(x=2\sqrt{2}\)vào a ra có
\(\sqrt{3-2\sqrt{2}}-2\sqrt{3}+1=\sqrt{\left(\sqrt{2}-1\right)^2}-2\sqrt{3}+1\)
\(=\sqrt{2}-1+2\sqrt{3}+1=\sqrt{2}+2\sqrt{3}\)
Để A=1<=> \(\sqrt{3-x}-2\sqrt{3}+1=1\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}+1-1=0\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}=0\\ \Leftrightarrow3-x=12\Leftrightarrow x=-9\)
a)
Xét hiệu \(\frac{a^3}{a^2+1}-\frac{1}{2}=\frac{2a^3-a^2-1}{2\left(a^2+1\right)}=\frac{2a^2\left(a-1\right)+\left(a-1\right)\left(a+1\right)}{2\left(a^2+1\right)}=\frac{\left(a-1\right)\left(2a^2+a+1\right)}{2\left(a^2+1\right)}\)
Do : \(a\ge1\Rightarrow a-1\ge0\)
\(a^2+a+1=\left(a+\frac{1}{4}\right)^2+\frac{3}{4}>0\Rightarrow2a^2+a+1>0\)
\(a^2+1>0\)
\(\Rightarrow\frac{\left(a-1\right)\left(2a^2+a+1\right)}{2\left(a^2+1\right)}\ge0\Leftrightarrow\frac{a^3}{a^2+1}-\frac{1}{2}\ge0\Leftrightarrow\frac{a^3}{a^2+1}\ge\frac{1}{2}\)
Tương tự \(\frac{b^3}{b^2+1}\ge\frac{1}{2};\frac{c^3}{c^2+1}\ge\frac{1}{2}\)
\(\Rightarrow\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1}\ge\frac{3}{2}\)Dấu = xảy ra khi a=b=c=1
\(\sqrt{\frac{3,6}{4,9}}=\pm\frac{6}{7}\)
\(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,664}\)
\(=\sqrt{1,08}\)
c tự tính đi
a) \(C=\frac{x-9}{x+6\sqrt{x}+9}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)^2}=\frac{\sqrt{x}-3}{\sqrt{x+3}}\)
b) Khi \(x=25\Rightarrow C=\frac{\sqrt{25}-3}{\sqrt{25}+3}=\frac{5-3}{5+3}=\frac{2}{8}=\frac{1}{4}\)
Khi \(x=4-2\sqrt{3}=3-2\sqrt{3}.1+1=\left(\sqrt{3}-1\right)^2\)
\(C=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}-3}{\sqrt{\left(\sqrt{3}-1\right)^2}+3}=\frac{\sqrt{3}-1-3}{\sqrt{3}-1+3}=\frac{\sqrt{3}-4}{\sqrt{3}+2}\)
c) Có \(\sqrt{x}\ge0\Rightarrow x\ge0\). Có \(\frac{\sqrt{x}-3}{\sqrt{x}+3}\) có \(\sqrt{x}+3>0\)
\(\Rightarrow min\left(\sqrt{x}-3\right)=min\frac{\sqrt{x}-3}{\sqrt{x}+3}=-3\Leftrightarrow x=0\)