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4:
a: góc CEH+góc CDH=180 độ
=>CDHE nội tiếp
b: Xét ΔHEA vuông tại E và ΔHDB vuông tại D có
góc EHA=góc DHB
=>ΔHEA đồng dạng với ΔHDB
=>HE/HD=HA/HB
=>HE*HB=HD*HA
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,\Leftrightarrow\left\{{}\begin{matrix}m+2=1\\m\ne2\end{matrix}\right.\Leftrightarrow m=-1\\ c,\text{PT giao Ox: }y=0\Leftrightarrow\left(m+2\right)x-m=0\\ \text{Thay }x=2\Leftrightarrow2m+4-m=0\\ \Leftrightarrow m=-4\\ d,\text{PT giao Ox và Oy: }\\ y=0\Leftrightarrow x=\dfrac{m}{m+2}\Leftrightarrow A\left(\dfrac{m}{m+2};0\right)\Leftrightarrow OA=\left|\dfrac{m}{m+2}\right|\\ x=0\Leftrightarrow y=-m\Leftrightarrow B\left(0;-m\right)\Leftrightarrow OB=\left|m\right|\\ \Delta OAB\text{ cân }\Leftrightarrow OA=OB\Leftrightarrow\left|\dfrac{m}{m+2}\right|=\left|m\right|\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{m}{m+2}=m\\\dfrac{m}{m+2}=-m\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m\left(m+1\right)=0\\m\left(m+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\\m=-3\end{matrix}\right.\)
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a: Xét tứ giác OBAC có
\(\widehat{OBA}+\widehat{OCA}=180^0\)
Do đó: OBAC là tứ giác nội tiếp
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Leftrightarrow x^2=\sqrt{4}\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)
b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)
\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)
c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)
\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)
f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)
\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)
\(\Leftrightarrow\sqrt{2-x}=4\)
\(\Leftrightarrow2-x=16\)
hay x=-14
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT cauchy, ta có:
\(\sqrt{\left(2y+2z-x\right)\cdot3x}\le\dfrac{2z+2y-x+3x}{2}=\dfrac{2\left(x+y+z\right)}{2}=x+y+z\\ \Leftrightarrow\sqrt{2y+2z-x}\le\dfrac{x+y+z}{\sqrt{3x}}\\ \Leftrightarrow\sqrt{\dfrac{x}{2y+2z-x}}\ge\dfrac{\sqrt{x}}{\dfrac{x+y+z}{\sqrt{3x}}}=\dfrac{x\sqrt{3}}{x+y+z}\)
\(\Leftrightarrow S=\sum\sqrt{\dfrac{x}{2y+2z-x}}\ge\sqrt{3}\left(\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\right)\\ \Leftrightarrow S\ge\sqrt{3}\cdot\dfrac{x+y+z}{x+y+z}=\sqrt{3}\)
Dấu \("="\Leftrightarrow x=y=z\) hay tam giác đều
![](https://rs.olm.vn/images/avt/0.png?1311)
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`Answer:`
`x^2+4x+7=(x+4)\sqrt{x^2+7}`
Ta đặt `v=\sqrt{x^2+7}` và `v>=\sqrt{7}`
`=>v^2=x^2+7`
Phương trình trở thành: `v^2+4x=(x+4)v`
`<=>v^2-xv+4x-4v=0`
`<=>(v-4)(v-x)=0`
`<=>v=4` hoặc `v=x`
Với `v=4` ta được: `\sqrt{x^2+7}=4`
`=>x^2+7=16`
`<=>x^2=9`
`<=>x=+-3`
Với `v=x` ta được: `\sqrt{x^2+7}=x`
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2+7=x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\7=0\text{(Vô lý)}\end{cases}}\Leftrightarrow x\in\varnothing\)
Đặt \(\sqrt{x^2+7}=t>0\)
Pt trở thành:
\(t^2-\left(x+4\right)t+4x=0\)
\(\Leftrightarrow t^2-xt-4t+4x=0\)
\(\Leftrightarrow t\left(t-x\right)-4\left(t-x\right)=0\)
\(\Leftrightarrow\left(t-x\right)\left(t-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=4\\\sqrt{x^2+7}=x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=16\\x^2+7=x^2\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm3\)