![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. Để biểu thức xác định thì:
$x^2-x-6\geq 0$
$\Leftrightarrow (x+2)(x-3)\geq 0$
$\Leftrightarrow x\geq 3$ hoặc $x\leq -2$
b. Để biểu thức xác định thì:
$4x-x^2-5\geq 0$
$\Leftrightarrow x^2-4x+5\leq 0$
$\Leftrightarrow (x-2)^2+1\leq 0$
$\Leftrightarrow (x-2)^2\leq -1< 0$ (vô lý)
Vậy không tồn tại $x$ để bt xác định
c. Để biểu thức xác định thì:
$x^2-8x+15>0$
$\Leftrightarrow (x-3)(x-5)>0$
$\Leftrightarrow x>5$ hoặc $x< 3$
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-2\end{matrix}\right.\)
b) ĐKXĐ: \(x\in\varnothing\)
c) ĐKXĐ: \(\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)ĐK:`4x^2-12x+9>0`
`<=>(2n-3)^2>0`
`<=>2n-3 ne 0`
`<=>n ne 3/2`
`d)x^2-x+1`
`=(x-1/2)^2+3/4>0AAx`
`=>` bt xd `AAx in RR`
e)ĐK:`x^2-8x+15>0`
`<=>x^2-3x-5x+15>0`
`<=>x(x-3)-5(x-3)>0`
`<=>(x-3)(x-5)>0`
`TH1:` \(\begin{cases}x-3>0\\x-5>0\\\end{cases}\)
`<=>` \(\begin{cases}x>3\\x>5\\\end{cases}\)
`<=>x>5`
`TH2:` \(\begin{cases}x-3<0\\x-5<0\\\end{cases}\)
`<=>` \(\begin{cases}x<3\\x<5\\\end{cases}\)
`<=>x<3`
f)ĐK:`3x^2-7x+20>0`
`<=>x^2-2x+1+2x^2-5x+19>0`
`<=>(x-1)^2+2(x-5/2)^2+13/2>0` luôn đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b) \(P=\sqrt{x}>3\Leftrightarrow x>9\)
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}\)
\(=\sqrt{x}\)
b: Để P>3 thì x>9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Leftrightarrow\Delta=64-4\left(2m+6\right)\ge0\\ \Leftrightarrow40-8m\ge0\\ \Leftrightarrow m\le5\\ 2,\Leftrightarrow\Delta=4\left(m-1\right)^2-4\left(2m-6\right)>0\\ \Leftrightarrow4m^2-8m+4-8m+24>0\\ \Leftrightarrow2\left(m^2-4m+4\right)+6>0\\ \Leftrightarrow2\left(m-2\right)^2+6>0\left(\text{luôn đúng}\right)\\ \Leftrightarrow m\in R\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{\sqrt{4x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b) Để P>3 thì \(\sqrt{x}>3\)
hay x>9
Kết hợp ĐKXĐ, ta được: x>9