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![](https://rs.olm.vn/images/avt/0.png?1311)
b: Để N là số nguyên dương thì \(\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
mà x là số nguyên
nên \(\left\{{}\begin{matrix}x\in Z\\x>9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x_1^2-1^2}+2\sqrt{x^2_2-2^2}+...+100\sqrt{x_{100}^2-100^2}=\dfrac{1}{2}\left(x_1^2+x^2_2+...+x_{100}^2\right)\)
\(\Leftrightarrow2\sqrt{x_1^2-1^2}+4\sqrt{x^2_2-2^2}+...+200\sqrt{x_{100}^2-100^2}=x_1^2+x^2_2+...+x_{100}^2\)
\(\Leftrightarrow x_1^2-1-2\sqrt{x_1^2-1}+1+x^2_2-4-4\sqrt{x^2_2-4}+4+...+x^2_{100}-10000-200\sqrt{x_{100}^2-10000}+10000=0\)
\(\Leftrightarrow\left(\sqrt{x^2_1-1}-1\right)^2+\left(\sqrt{x^2_2-4}-2\right)^2+....+\left(\sqrt{x^2_{100}-10000}-100\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2_1-1}-1=0\\\sqrt{x^2_2-4}-2=0\\....\\\sqrt{x^2_{100}-10000}-100=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\sqrt{1^2+1}=\sqrt{2}\\x_2=\sqrt{2^2+4}=2\sqrt{2}\\....\\x_{100}=\sqrt{100^2+10000}=100\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(\sqrt{\left(1+\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=a+b\sqrt{2}+c\sqrt{3}\)
Suy ra \(a=2;b=c=0\)
Cái này phá căn dùng đồng nhất thức nhé cái này mình làm bên học 24 rồi ko muốn làm lại
\(a^2+b^2+c^2=2^2+0+0=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
Ta có : \(\dfrac{3x+5}{2}-1\le\dfrac{x+2}{3}+x\)
\(\Leftrightarrow\dfrac{3x+5}{2}-1-\dfrac{x+2}{3}-x\le0\)
\(\Leftrightarrow\dfrac{3\left(3x+5\right)-6-2\left(x+2\right)-6x}{6}\le0\)
\(\Leftrightarrow9x+15-6-2x-4-6x\le0\)
\(\Leftrightarrow x\le-5\)
Mà \(\left\{{}\begin{matrix}x\in Z\\x>-10\end{matrix}\right.\)
Vậy \(x\in\left\{-5;-6;-7;-8;-9\right\}\)
b3\(\Leftrightarrow2x^2+5x-3-3x+1\le x^2+2x-3+x^2-5\\ \Leftrightarrow0.x\le-6\Leftrightarrow x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. BĐT tương đương với \(6\left(a^2+b^2\right)-2ab+8-4\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge0\)
\(\Leftrightarrow\left[a^2-4a\sqrt{b^2+1}+4\left(b^2+1\right)\right]+\left[b^2-4b\sqrt{a^2+1}+4\left(a^2+1\right)\right]\)\(+\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-2\sqrt{b^2+1}\right)^2+\left(b-2\sqrt{a^2+1}\right)^2+\left(a-b\right)^2\ge0\)(đúng)
=> Đẳng thức không xảy ra
2. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)
\(\Leftrightarrow a^4+b^4+c^2+1\ge2a^2b^2-2a^2+2ac+2a\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^2-2ac+a^2\right)+\left(a^2-2a+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c-a\right)^2+\left(a-1\right)^2\ge0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x=a+\sqrt{15};y=\dfrac{1}{a}-\sqrt{15}\left(x,y\in Z\right)\)
Ta có: \(y=\dfrac{1}{x-\sqrt{15}}-\sqrt{15}\Leftrightarrow xy-16=\left(y+x\right)\sqrt{15}\)
Nếu y=x thì VP là số vô tỉ còn VT là số nguyên ( vô lý)
=> x=y
=> xy-16=0 <=> x=y=\(\pm\)4 .
Thay vào tìm đc \(\left[{}\begin{matrix}a=4-\sqrt{15}\\a=-4-\sqrt{15}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=2\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2\)
\(B=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{6}-\sqrt{2}\right)\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{a\left(\sqrt{a}-1\right)\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{6}-\sqrt{2}}{a+\sqrt{ab}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)