![](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)
ĐKXĐ: \(\left\{{}\begin{matrix}5x^2+14x+9>=0\\x+1>=0\\x^2-x-20>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(5x+9\right)>=0\\x+1>=0\\\left(x-5\right)\left(x+4\right)>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< =-\dfrac{9}{5}\\x>=-1\end{matrix}\right.\\x>=-1\\\left[{}\begin{matrix}x>=5\\x< =-4\end{matrix}\right.\end{matrix}\right.\)
=>x>=5
\(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
=>\(\sqrt{5x^2+14x+9}-21+6-\sqrt{x^2-x-20}=5\sqrt{x+1}-15\)
=>\(\dfrac{5x^2+14x+9-441}{\sqrt{5x^2+14x+9}+21}+\dfrac{36-x^2+x+20}{6+\sqrt{x^2-x-20}}=5\left(\sqrt{x+1}-3\right)\)
=>\(\dfrac{5x^2+14x-432}{\sqrt{5x^2+14x+9}+21}+\dfrac{-x^2+x+56}{6+\sqrt{x^2-x-20}}=5\cdot\dfrac{x+1-9}{\sqrt{x+1}+3}\)
=>\(\dfrac{\left(x-8\right)\left(5x+54\right)}{\sqrt{5x^2+14x+9}+21}-\dfrac{x^2-x-56}{\sqrt{x^2-x-20}+6}=\dfrac{5\left(x-8\right)}{\sqrt{x+1}+3}\)
=>\(\dfrac{\left(x-8\right)\left(5x+4\right)}{\sqrt{5x^2+14x+9}+21}-\dfrac{\left(x-8\right)\left(x+7\right)}{\sqrt{x^2-x-20}+6}-\dfrac{5\left(x-8\right)}{\sqrt{x+1}+3}=0\)
=>\(\left(x-8\right)\left(\dfrac{5x+4}{\sqrt{5x^2+14x+9}+21}-\dfrac{x+7}{\sqrt{x^2-x-20}+6}-\dfrac{5}{\sqrt{x+1}+3}\right)=0\)
=>x-8=0
=>x=8(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2
a, bạn tự vẽ
b, Hoành độ giao điểm tm pt
\(2x^2-2x+3=0\)
\(\Delta'=1-3.2=-5< 0\)
Vậy pt vô nghiệm hay (d) ko cắt (P)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ M=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{-4\sqrt{a}}{a-1}=\dfrac{1-a}{\sqrt{a}}\)
anh có thể ghi thêm các bước trước khi ra đc mấy cái này ko ạ tại rút gọn quá e ch hỉu ạ e c.ơn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,m=0\Leftrightarrow y=3x+2\)
Vì \(3>0\) nên hàm đồng biến
\(b,\text{Thay }x=-1;y=3\\ \Leftrightarrow-m-3+2=3\Leftrightarrow m=-4\\ c,\text{PT giao Ox: }y=0\Leftrightarrow x=-\dfrac{2}{m+3}\Leftrightarrow A\left(-\dfrac{2}{m+3};0\right)\Leftrightarrow OA=\dfrac{2}{\left|m+3\right|}\\ \text{PT giao Oy: }x=0\Leftrightarrow y=2\Leftrightarrow B\left(0;2\right)\Leftrightarrow OB=2\\ \text{Ta có }S_{OAB}=4\\ \Leftrightarrow\dfrac{1}{2}OA\cdot OB=4\Leftrightarrow\dfrac{2}{\left|m+3\right|}\cdot2=8\\ \Leftrightarrow\dfrac{4}{\left|m+3\right|}=8\\ \Leftrightarrow\left|m+3\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{5}{2}\\m=-\dfrac{7}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{5}.\sqrt{7}}}.\left(3\sqrt{2}+\sqrt{2}.\sqrt{7}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\sqrt{\dfrac{1}{8+3\sqrt{7}}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\dfrac{\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{8+3\sqrt{7}}}\)
\(=\dfrac{\sqrt{2}.\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{2}.\sqrt{8+3\sqrt{7}}}\) (Nhân \(\sqrt{2}\) cả tử và mẫu)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{16+6\sqrt{7}}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{\left(3+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\left|3+\sqrt{7}\right|}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{3+\sqrt{7}}\)
\(=2=VP\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`A=1/(x+sqrtx)+(2sqrtx)/(x-1)-1/(x-sqrtx)`
`=(sqrtx-1+2x-sqrtx-1)/(sqrtx(x-1))`
`=(2x-2)/(sqrtx(x-1))`
`=2/sqrtx`
`b)A=1`
`<=>2/sqrtx=1`
`<=>sqrtx=2`
`<=>x=4(tm)`
nhiều quá không viết được