Tìm GTLN của:
Q = \(\dfrac{3\left(x+1\right)}{x^{3^{ }}+x^2+x+1}\)
help me please❤⚽☘
K
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DT
Dang Tung
CTVHS
18 tháng 10 2023
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)
H
18 tháng 10 2023
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)
30 tháng 6 2017
a, Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu " = " khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
b, Để B lớn nhất thì \(\left(x-\dfrac{2}{3}\right)^2+9\) nhỏ nhất
Ta có: \(\left(x-\dfrac{2}{3}\right)^2+9\ge9\)
\(\Leftrightarrow B=\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\le\dfrac{4}{9}\)
Dấu " = " khi \(\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(MAX_B=\dfrac{4}{9}\) khi \(x=\dfrac{2}{3}\)
RF
2
FC
22 tháng 11 2018
Ta có:
\(\dfrac{\left(x-1\right)^2}{9}=\dfrac{3}{x-1}\)
=> (x - 1)2 . (x - 1) = 9 . 3
(x - 1)3 = 27
=> \(x=\sqrt[3]{27}+1=3+1=4\)
Vậy x = 4
please
13 tháng 11 2017
\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)
HN
10 tháng 3 2017
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)
ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)
\(\Leftrightarrow3x^2+15x-378=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)
10 tháng 3 2017
@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T
27 tháng 8 2017
Do mỗi số hạng ở vế trái nằm trong dấu giá trị tuyệt đối mà vế phải 100 là số dương nên x cũng là số dương
Do x dương nên ta có:
\(x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+...+x+\dfrac{1}{99.100}=100x\)
Dãy trên có 99 số hạng nên
\(99x+\left(x-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(1-\dfrac{1}{100}=x\Rightarrow x=\dfrac{99}{100}\)
Vậy \(x=\dfrac{99}{100}\)
NM
3
10 tháng 8 2023
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Q = \(\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)để Q đạt GTLN => \(x^2+1phảiNhỏnhất\)
\(x^2+1\ge1=>x^2+1\)đạt GTNN là 1 khi x=0
vậy Q đạt GTLN =3 khi x = 0
\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}\\ Q=\dfrac{3}{x^2+1}\\ Do\text{ }x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\\ Q=\dfrac{3}{x^2+1}\le3\forall x\\ \text{Dấu “=” xảy ra khi : }\\ x^2=0\\ \Leftrightarrow x=0\\Vậy\text{ }Q_{\left(Max\right)}=3\text{ }khi\text{ }x=0\)