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BC
4
BC
0
BC
3
27 tháng 9 2021
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)
Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)
\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)
PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)
+ Với a=1
\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)
+ Với b=1
\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
HN
27 tháng 9 2021
Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)
Thì được:
\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)
Làm tiếp
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
21 tháng 3 2016
<=><=>(X+1)(Y+1)=6 và (x+1)^3+(y+1)^3=35đặt X+1;Y+1 biến đổi vế 2 giải ra đc(1;2);(2;1)
b,<=>\(\left[\sqrt{2}+1\right]^x+\left[\sqrt{2}-1\right]^x=6\)
<=>\(2\sqrt{2}^x+2=6\)
<=>x=2
A
0
\(\sqrt{\left(x-1\right)\left(x-5\right)}-3\sqrt{1-x}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-5\right)\ge0\\1-x\ge0\\\sqrt{\text{}\left(x-1\right)\left(x-5\right)}=3\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5\ge0\\x-1\le0\end{matrix}\right.\\x\le1\\\sqrt{\text{}\left(x-1\right)\left(x-5\right)}=3\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\\x\le1\\\left(x-1\right)\left(x-5\right)=9\left(1-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-5\right)+9\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-5+9\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
- Vậy \(S=\left\{1;-4\right\}\)
\(\sqrt{1-x}(\sqrt{5-x}-3)=0\)
x = 1 hay \(\sqrt{5-x}=3 \)
x = 1 hay x= - 4