có ai giúp em câu này với ạ em cảm ơn.
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 12 2021
\(b,\Leftrightarrow\left\{{}\begin{matrix}m+1=3\\m-3\ne-3\end{matrix}\right.\Leftrightarrow m=2\\ c,\text{PT giao Ox tại hoành độ 3: }\\ x=-3;y=0\Leftrightarrow\left(m+1\right)\left(-3\right)+m-3=0\\ \Leftrightarrow-2m-6=0\Leftrightarrow m=-3\)
PK
3
MH
19 tháng 2 2022
\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)
\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)
\(\Leftrightarrow2x+8=-x+11\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
23 tháng 1 2022
a, thay x=25 vào A ta có:
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{25}}{\sqrt{25}-1}=\dfrac{5}{5-1}=\dfrac{5}{4}\)
b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\dfrac{3x+3}{x\sqrt{x}-1}-\dfrac{2}{\sqrt{x}-1}\right)\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\dfrac{3x+3}{\sqrt{x^3}-1}-\dfrac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\dfrac{3x+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{2x+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{x}-1}.\dfrac{3x+3-2x-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
22 tháng 9 2021
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{5\sqrt{3}}{2}-9\sqrt{3}=\dfrac{5\sqrt{3}-18\sqrt{3}}{2}=\dfrac{-13\sqrt{3}}{2}\)
22 tháng 9 2021
\(=\dfrac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5.\dfrac{\sqrt{3}}{2}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{5\sqrt{3}}{2}\)
\(=-9\sqrt{3}+\dfrac{5\sqrt{3}}{2}=\dfrac{-18\sqrt{3}+5\sqrt{3}}{2}=-\dfrac{13\sqrt{3}}{2}\)
N
23 tháng 1 2022
Giải hpt:
Đặt: \(\left[{}\begin{matrix}\sqrt{x-1}=a\\y+1=b\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}3a-2b=-1\\5a-9b=-13\end{matrix}\right.< =>\left\{{}\begin{matrix}15a-10b=-5\\15a-27b=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\15a-27\cdot2=-39\end{matrix}\right.< =>\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\)
Thay: \(\left[{}\begin{matrix}\sqrt{x-1}=1\\y+1=2\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2022
\(S=\dfrac{1}{2^2}+\dfrac{1}{\left(2.2\right)^2}+\dfrac{1}{\left(2.3\right)^2}+...+\dfrac{1}{\left(2.10\right)^2}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2^2.2^2}+\dfrac{1}{2^2.3^2}+...+\dfrac{1}{2^2.10^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\right)\)
\(< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\)
\(=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{4}\left(2-\dfrac{1}{10}\right)< \dfrac{1}{4}.2=\dfrac{1}{2}\) (đpcm)
SE
1
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 4 2021
\(A=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(c+1\right)}+\sqrt{2c\left(a+1\right)}\)
\(A=\dfrac{1}{\sqrt{2}}\sqrt{4a\left(b+1\right)}+\dfrac{1}{\sqrt{2}}\sqrt{4b\left(c+1\right)}+\dfrac{1}{\sqrt{2}}\sqrt{4c\left(a+1\right)}\)
\(A\le\dfrac{1}{2\sqrt{2}}\left(4a+b+1\right)+\dfrac{1}{2\sqrt{2}}\left(4b+c+1\right)+\dfrac{1}{2\sqrt{2}}\left(4c+a+1\right)\)
\(A\le\dfrac{1}{2\sqrt{2}}\left[5\left(a+b+c\right)+3\right]=2\sqrt{2}\)
\(A_{max}=2\sqrt{2}\) khi \(a=b=c=\dfrac{1}{3}\)
2 tháng 12 2021
\(1,\left\{{}\begin{matrix}3x-y=5\\5x+2y=23\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\5x+2\left(3x-5\right)=23\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\5x+6x-10=23\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3.3-5\\x=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
\(2,\left\{{}\begin{matrix}5x-4y=3\\2x+y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-4\left(4-2x\right)=3\\y=4-2x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-16+8x=3\\y=4-2x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}13x=19\\y=4-2x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{13}\\y=4-2.\dfrac{19}{13}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{13}\\y=\dfrac{14}{13}\end{matrix}\right.\)
