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\(=\frac{3\left(x^2+2x+3\right)+1}{\left(x^2+2x+3\right)}=3+\frac{1}{\left(x+1\right)^2+2}\). ta có: \(\left(x+1\right)^2\ge0\Leftrightarrow\left(x+1\right)^2+2\ge2\Leftrightarrow\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\Leftrightarrow3+\frac{1}{\left(x+1\right)^2+2}\le\frac{7}{2}\)
=> max M=7/2 <=> x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)
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\(a,A=\left(1;2\right)\Leftrightarrow x=1;y=2\\ \Leftrightarrow2=\left(m+1\right)-2m+3\\ \Leftrightarrow-m+4=2\Leftrightarrow m=2\)
\(c,\)Giả sử điểm cố định là \(A\left(x_0;y_0\right)\)
\(\Leftrightarrow y_0=\left(m+1\right)x_0-2m+3\\ \Leftrightarrow y_0=mx_0+x_0-2m+3\\ \Leftrightarrow m\left(x_0-2\right)+\left(x_0-y_0+3\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0-2=0\\x_0-y_0+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=2\\y_0=5\end{matrix}\right.\Leftrightarrow B\left(2;5\right)\)
Vậy \(\left(d\right)\) luôn đi qua điểm \(B\left(2;5\right)\) cố định
\(d,\) Pt hoành độ giao điểm:
\(2=\left(2+1\right)x-2\cdot2+3\\ \Leftrightarrow2=3x-1\Leftrightarrow x=1\\ \Leftrightarrow C\left(1;2\right)\)
Vậy ...
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(H=\left(\sqrt{4x^2-12x+9}+\sqrt{4x^2+4x+1}\right)\)
\(\Leftrightarrow H=\left(\sqrt{\left(2x-3\right)^2}+\sqrt{\left(2x+1\right)^2}\right)\)
\(\Leftrightarrow H=\left|2x-3\right|+\left|2x+1\right|\)
Xét tính chất về trị tuyệt đối sau: \(\left|a\right|+\left|b\right|\ge ab\) với \(ab\ge0\)
Ta viết lại \(H=\left|3-2x\right|+\left|2x+1\right|\ge\left|\left(3-2x\right)+\left(2x+1\right)\right|=4\) khi \(\left(3-2x\right)\left(2x+1\right)\ge0\)
\(\Rightarrow H\ge4\)khi \(3-2x\ge0\)và \(2x+1\ge0\) hoặc \(3-2x\le0\) và \(2x+1\le0\)
\(\Leftrightarrow x\le\frac{3}{2}\) và \(x\ge\frac{-1}{2}\)hoặc \(x\ge\frac{3}{2}\)và \(x\le\frac{-1}{2}\)(vô lý)
Vậy \(GTNN\left(H\right)=4\) khi \(\frac{-1}{2}\le x\le\frac{3}{2}\)
Mình có giải thích hơi dài nha cậu tick mình nha
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2.7
a/ $9+4\sqrt{5}=2^2+2.2\sqrt{5}+(\sqrt{5})^2=(2+\sqrt{5})^2$
b/ $\sqrt{9+4\sqrt{5}}-\sqrt{5}=\sqrt{(2+\sqrt{5})^2}-\sqrt{5}$
$=|2+\sqrt{5}|-\sqrt{5}=2+\sqrt{5}-\sqrt{5}=2$
c/ $\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{4^2+2.4\sqrt{7}+(\sqrt{7})^2}-\sqrt{7}=\sqrt{(4+\sqrt{7})^2}-\sqrt{7}$
$=4+\sqrt{7}-\sqrt{7}=4$
d.
$\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}$
$=\sqrt{(a-2)+2.2\sqrt{a-2}+2^2}+\sqrt{(a-2)-2.2\sqrt{a-2}+2^2}$
$=\sqrt{(\sqrt{a-2}+2)^2}+\sqrt{(\sqrt{a-2}-2)^2}$
$=|\sqrt{a-2}+2|+|\sqrt{a-2}-2|$
$=\sqrt{a-2}+2+2-\sqrt{a-2}=4$ (do $a\leq 6$ nên $\sqrt{a-2}-2\leq 0$ nên $|\sqrt{a-2}-2|=2-\sqrt{a-2}$)
2.5
a.
$\sqrt{(x-3)^2}=3-x$
$\Leftrightarrow |x-3|=3-x$
$\Leftrightarrow 3-x\geq 0$
$\Leftrightarrow x\leq 3$
b.
$\sqrt{25-20x+4x^2}+2x=5$
$\Leftrightarrow \sqrt{(2x-5)^2}=5-2x$
$\Leftrightarrow |2x-5|=5-2x$
$\Leftrightarrow 5-2x\geq 0$
$\Leftrightarrow x\leq \frac{2}{5}$
c.
$\sqrt{x^2-\frac{1}{2}x+\frac{1}{16}}=\frac{1}{4}-x$
$\Leftrightarrow \sqrt{(x-\frac{1}{4})^2}=\frac{1}{4}-x$
$\Leftrightarrow |x-\frac{1}{4}|=\frac{1}{4}-x$
$\Leftrightarrow \frac{1}{4}-x\geq 0$
$\Leftrightarrow x\leq \frac{1}{4}$
\(b,=5\sqrt{2}-3\sqrt{3}-\sqrt{\left(5\sqrt{2}-3\right)^2}=5\sqrt{2}-3\sqrt{3}-5\sqrt{2}+3=3-3\sqrt{3}\)