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\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)
\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)
\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất
Mà \(\left|2018x-2019\right|\ge0\)
\(\Rightarrow\left|2018x-2019\right|+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left|2018x-2019\right|=0\)
\(\Leftrightarrow x=\frac{2019}{2018}\)
Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)
\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)
\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)
\(\Rightarrow5^x=3^{2x}\)
Mà \(\left(5;3\right)=1\)
\(\Rightarrow x=2x=0\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
a)Đang suy nghĩ...
b)\(M\left(x\right)=\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
a) \(12x^{11}-15x^7-6x^5+2018\)
\(=3x^5.\left(4x^6-5x^2-2\right)+2018\)
\(=3x^5.0+2018\)
\(=2018\)
1. A = 100
2. B = 2098
mik ko biết có đúng ko đâu nhé vì mình nhowfbanj làm cho rồi viết vô đây mà ahihi
a) \(\begin{cases}\left(x+2\right)^2\ge0\\\left(y-\frac{1}{5}\right)^2\ge0\end{cases}\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10=-10\)hay \(C\ge-10\)
Dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-\frac{1}{5}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{5}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}}}\)
Vậy GTNN C là -10 khi \(\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}.}\)
b)\(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+5\ge0+5=5\)
\(\Rightarrow\frac{4}{\left(2x-3\right)^2-5}\le\frac{4}{5}\Leftrightarrow D\le\frac{4}{5}\)
Dấu "=" xảy ra khi:
\(\left(2x-3\right)^2=0\Rightarrow2x-3=0\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)
Vậy GTLN D là \(\frac{4}{5}\)khi \(x=\frac{3}{2}.\)
1. Tìm x
\(3x.\left(2x-\frac{3}{5}\right)=0\)
⇒ \(\left[{}\begin{matrix}3x=0\\2x-\frac{3}{5}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=0:3\\2x=0+\frac{3}{5}=\frac{3}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=0\\x=\frac{3}{5}:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=0\\x=\frac{3}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\frac{3}{10}\right\}.\)
2.
b)
TH1: \(a< b.\)
\(\Rightarrow2019a< 2019b\)
\(\Rightarrow ab+2019a< ab+2019b.\)
\(\Rightarrow a.\left(b+2019\right)< b.\left(a+2019\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+2019}{b+2019}.\)
TH2: \(a=b.\)
\(\Rightarrow\frac{a}{b}=\frac{a+2019}{b+2019}.\)
TH3: \(a>b.\)
\(\Rightarrow2019a>2019b.\)
\(\Rightarrow ab+2019a>ab+2019b\)
\(\Rightarrow a.\left(b+2019\right)>b.\left(a+2019\right).\)
\(\Rightarrow\frac{a}{b}>\frac{a+2019}{b+2019}.\)
Chúc bạn học tốt!
1
x . (2x - 3/ 5 ) = 0 : 3
2x - 3/ 5 = 0
2x = 0+3/5
x =3/ 5 : 2
x = 3/ 10