tìm x
(x+3/4).(x+2/3)<= 0
x^2= 3/4x<0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có
\(C=\left(3-x\right)\left(1-y\right)\left(4x-7y\right)\)
\(\Leftrightarrow28C=\left(12-4x\right)\left(7-7y\right)\left(4x-7y\right)\)
\(\Leftrightarrow3.\sqrt[3]{28C}=3.\sqrt[3]{\left(12-4x\right)\left(7-7y\right)\left(4x-7y\right)}\)
\(\le12-4x+7-7y+4x-7y=19\)
\(\Leftrightarrow\sqrt[3]{28C}\le\frac{19}{3}\)
\(\Leftrightarrow28C\le\frac{19^3}{27}\)
\(\Leftrightarrow C\le\frac{19^3}{27.28}\)
\(\left(2x-4\right)\left(1-3x\right)=0\)
<=> \(2\left(x-2\right)\left(1-3x\right)=0\)
<=> \(\orbr{\begin{cases}x-2=0\\1-3x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}\)
Vậy....
\(\left(2x-4\right)\left(1-3x\right)=0\)
\(\Rightarrow2x-4=0\)hoặc\(1-3x=0\)
\(TH1:2x-4=0\)
\(2x=0+4\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
\(TH2:1-3x=0\)
\(3x=1-0\)
\(3x=1\)
\(x=\frac{1}{3}\)
Vậy:\(x=2\)hoặc \(x=\frac{1}{3}\)
Câu 1
\(x^3-2x^2+3x-6< 0\\ \Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)< 0\\ \Leftrightarrow\left(x-2\right)\left(x^2+3\right)< 0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2< 0\Leftrightarrow x>2\\x^2+3< 0\Leftrightarrow x^2< 0\Leftrightarrow x\in\varnothing\end{matrix}\right.\)
S = {x/x>2}
câu 1 : tách 6=2.3
Câu 2: tách -4x = -3x-x
Câu 3 tách x= 2x-3x
Bài 3:
\(\Leftrightarrow3^{2x+6}=3\)
=>2x+6=1
=>2x=-5
hay x=-5/2
1. 4x2 + 4x + 2 = (4x2 + 4x + 1) + 1 = (2x + 1)2 + 1
Có: (2x+1)2 ≥ 0 ∀x => (2x+1)2 + 1 ≥ 1 > 0 (đpcm)
3. -x2 + 4x - 5 = -(x2 - 4x + 4) - 1 = -(x - 2)^2 - 1
Có: -(x-2)^2 ≤ 0 => -(x-2)^2 -1 ≤ - 1 < 0 (đpcm)
7. (x+2)(x-5) + 15 = x2 - 3x + 5 = (x2 - 2.x.\(\dfrac{3}{2}\)+ \(\dfrac{9}{4}\)) + \(\dfrac{11}{4}\)
= ( x - \(\dfrac{3}{2}\))^2 + \(\dfrac{11}{4}\) \(\ge\dfrac{11}{4}>0\left(đpcm\right)\)