5+4x-x+2=(5+4x).(7+5x)
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(x-3).(2x-1)=(2x-1).(2x+3)
<=> (x-3).(2x-1)-(2x-1).(2x+3)=0
<=> (x-3-2x-3)(2x-1)=0
<=> (-3x-6)(2x-1)=0
<=> -3x-6=0 hoặc 2x-1=0
<=> -3x=6 hoặc 2x=1
<=> x=-2 hoặc x=1/2
Vậy \(x\in\left\{-2;\frac{1}{2}\right\}\)
(x - 3)(2x - 1) = (2x - 1)(2x + 3)
<=> (x - 3)(2x - 1) - (2x - 1)(2x + 3) = 0
<=> (2x - 1)(x - 3 - 2x - 3) = 0
<=> (2x - 1)(-x - 6) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\-x-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-6\end{cases}}\)
Vậy S = {1/2; -6}
5x2 + 8xy + 5y2 = 72
<=> 5x2 + 10xy + 5y2 - 2xy = 72
<=> 5(x2 + 2xy + y2) - 2xy = 72
<=> 5(x + y)2 - 2xy = 72
<=> -2xy = 72 - 5(x + y)2
A = x2 + y2 = (x + y)2 - 2xy
= (x + y)2 + 72 - 5(x + y)2
= 72 - 4(x + y)2
(x + y)2 > 0 => -4(x + y)2 < 0
=> A < 72
dấu "=" xảy ra khi : x + y = 0 <=> x = -y
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, Theo ĐLBTKL ta có :
\(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)
có \(m_{Mg}=24g;m_{HCl}=36,5g;m_{H_2}=2g\)
\(\Rightarrow m_{MgHCl=24+36,5-2=58,5g}\)
\(5+4x-x+2=\left(5x+4\right)\left(7+5x\right)\)
\(\Leftrightarrow5+4x-x+2=35+28x+25x+20x^2\)
\(\Leftrightarrow x^2+50x+28=0\)
Ta có \(\Delta=50^2-4.1.28=2388,\sqrt{\Delta}=2\sqrt{597}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-50+2\sqrt{597}}{2}=-25+\sqrt{597}\\x=\frac{-50-2\sqrt{597}}{2}=-25-\sqrt{597}\end{cases}}\)
\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)
\(7+3x=\left(5+4x\right)\left(7+5x\right)\)
\(7+3x=35+28x+25x+20x^2\)
\(7+3x-35-28x-25x-20x^2=0\)
\(-28-50x-20x^2=0\)
\(-28-50x-20x^2=0\)
\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)