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d.
ĐKXĐ: \(x\left|x\right|-4>0\)
\(\Leftrightarrow x\left|x\right|>4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x^2>4\end{matrix}\right.\) \(\Leftrightarrow x>2\)
e.
ĐKXĐ: \(\left|x^2-2x\right|+\left|x-1\right|\ne0\)
Ta có:
\(\left|x^2-2x\right|+\left|x-1\right|=0\Leftrightarrow\left\{{}\begin{matrix}x^2-2x=0\\x-1=0\end{matrix}\right.\) (ko tồn tại x thỏa mãn)
\(\Rightarrow\) Hàm xác định với mọi x hay \(D=R\)
f.
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ge0\\x\left|x\right|+4\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\left|x\right|+4\ne0\end{matrix}\right.\)
Xét \(x\left|x\right|+4=0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x^2+4=0\left(vn\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\-x^2+4=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=-2\)
Hay \(x\left|x\right|+4\ne0\Leftrightarrow x\ne-2\)
Kết hợp với \(x\ge-2\Rightarrow x>-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \({\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4.{\left( {2x} \right)^3}{.1^1} + 6.{\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4} = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1\)
b) \(\begin{array}{l}{\left( {3y - 4} \right)^4} = {\left[ {3y + \left( { - 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4.{\left( {3y} \right)^3}.\left( { - 4} \right) + 6.{\left( {3y} \right)^2}.{\left( { - 4} \right)^2} + 4.{\left( {3y} \right)^1}{\left( { - 4} \right)^3} + {\left( { - 4} \right)^4}\\ = 81{y^4} - 432{y^3} + 864{y^2} - 768y + 256\end{array}\)
c) \({\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4.{x^3}.{\left( {\frac{1}{2}} \right)^1} + 6.{x^2}.{\left( {\frac{1}{2}} \right)^2} + 4.x.{\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}\)
d) \(\begin{array}{l}{\left( {x - \frac{1}{3}} \right)^4} = {\left[ {x + \left( { - \frac{1}{3}} \right)} \right]^4} = {x^4} + 4.{x^3}.{\left( { - \frac{1}{3}} \right)^1} + 6.{x^2}.{\left( { - \frac{1}{3}} \right)^2} + 4.x.{\left( { - \frac{1}{3}} \right)^3} + {\left( { - \frac{1}{3}} \right)^4}\\ = {x^4} - \frac{4}{3}{x^3} + \frac{2}{3}{x^2} - \frac{4}{27}x + \frac{1}{{81}}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=x^2-4\\x^2-5x-4=4-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x=0\\2x^2-5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5\pm\sqrt{89}}{4}\\\end{matrix}\right.\)
b/ - Với \(x\ge3\) pt trở thành:
\(x-1+3\left(x-3\right)=6\Leftrightarrow4x=16\Rightarrow x=4\)
- Với \(x\le1\) pt trở thành:
\(1-x+3\left(3-x\right)=6\)
\(\Leftrightarrow x=1\)
- Với \(1< x< 3\) pt trở thành:
\(x-1+3\left(3-x\right)=6\)
\(\Leftrightarrow-2x=-2\Rightarrow x=1\) (loại)
c/ ĐKXĐ: \(x\ne\pm2\)
\(\left[{}\begin{matrix}\frac{x^2-6x-4}{x^2-4}=1\\\frac{x^2-6x-4}{x^2-4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-4=x^2-4\\x^2-6x-4=4-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-6x=0\\2x^2-6x-8=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=4\end{matrix}\right.\)
d/ - Với \(x\ge2\) pt trở thành:
\(x-1-2\left(x-2\right)=x^2-x-3\)
\(\Leftrightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\left(l\right)\end{matrix}\right.\)
- Với \(x\le1\) pt trở thành:
\(1-x-2\left(2-x\right)=x^2-x-3\) làm tương tự
- Với \(1< x< 2\):
\(x-1-2\left(2-x\right)=x^2-x-3\)