15+4x<2x-145 , -3*(2x+5) - 16<-4*(3 - 2x)
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\(\left(4x-15\right)^{2016}=\left(4x-15\right)^{2015}\\ \Leftrightarrow\left[{}\begin{matrix}4x-15=0\\4x-15=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=15\\4x=16\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{15}{4};4\right\}\)
\(\left(4.x-15\right)^{2016}=\left(4.x-15\right)^{2015}\)
=>\(\left(4.x-15\right)^{2016}-\left(4.x-15\right)^{2015}=0\)
=>\(\left(4.x-15\right)^{2015}.\left[\left(4.x-15\right)-1\right]=0\)
=>\(\hept{\begin{cases}\left(4.x-15\right)^{2015}=0\\\left(4.x-15\right)-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x-15=0\\4.x-15-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x=15\\4x=16\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)
Vậy.................
(4x - 15)2016 = (4x - 15)2015
(4x - 5)2016 - (4x - 15)2015 = 0
(4x - 5)2015.(4x - 15) - (4x - 15)2015 = 0
(4x - 5)2015.[(4x - 15) - 1] = 0
=> \(\orbr{\begin{cases}\left(4x-5\right)^{2015}=0\\\left(4x-15\right)-1=0\end{cases}}\)=>\(\orbr{\begin{cases}4x-15=0\\4x-15=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=15\\4x=16\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)
tìm x,biết
5-x-16=40+x
4x-10=15-x
15-x=4x-5
x-15=6+4x
-12+x=5x-20
7x-4=20+3x
5x-7=-21-2x
x+15=20-4x
17-x=7-6x
Ta có : \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left(2x-1\right)^2-4\le-4\)
\(\Rightarrow B\ge\dfrac{15}{-4}\)
Dấu ''='' xảy ra khi x = 1/2
Vậy GTNN B là -15/4 khi x = 1/2
Đặt \(x^2-4x=t\)
\(\Rightarrow t^2-8t+15=0\)
\(\Leftrightarrow t^2-3t-5t+15=0\)
\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)
\(\Leftrightarrow\left(t-3\right)\left(t-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=5\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4x=5\\x^2-4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=0\\x^2-4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-5x-5=0\\x^2-4x+4-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)-5\left(x+1\right)=0\\\left(x-2\right)^2-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(x-5\right)=0\\\left(x-2-\sqrt{7}\right)\left(x-2+\sqrt{7}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\\x=2+\sqrt{7}\\x=2-\sqrt{7}\end{matrix}\right.\)
6.4\(x\).41 + 4\(^x\).4\(^3\) = 28.(7 + 15)
4\(^x\).(6.4 + 43) = 28.22
4\(^x\).(24 + 64) = 256.22
4\(^x\). 88 = 5632
4\(^x\) = 532 : 88
4\(^x\) = 64
4\(^x\) = 43
\(x\) = 3
\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\\ \Leftrightarrow4x^2+x-12x-3-\left(4x^2-28x-x+7\right)-15=0\\ \Leftrightarrow4x^2-11x-3-4x^2+29x-7-15=0\\ \Leftrightarrow18x=25\\ \Leftrightarrow x=\dfrac{25}{18}\)
Vậy \(x=\dfrac{25}{18}\)
Lời giải:
$(4x+1)(x-3)-(x-7)(4x-1)=15$
$\Leftrightarrow (4x^2-11x-3)-(4x^2-29x+7)=15$
$\Leftrightarrow 18x-10=15$
$\Leftrightarrow 18x=25$
$\Leftrightarrow x=\frac{25}{18}$
\(\Rightarrow15+4x-2x+145< 0\)
\(\Rightarrow\left(15+145\right)+\left(4x-2x\right)< 0\)
\(\Rightarrow160+2x< 0\)
\(\Rightarrow2x< 0-160\)
\(\Rightarrow2x< -160\)
\(\Rightarrow x< \left(-160\right)\div2\)
\(\Rightarrow x< -80\)
Vậy x < -80
\(\Rightarrow\left(-3\right)\times2x+\left(-3\right)\times5-16< \left(-4\right)\times3-\left(-4\right)\times2x\)
\(\Rightarrow-6x+\left(-15\right)-16< \left(-12\right)-\left(-8x\right)\)
\(\Rightarrow-6x-15-16< \left(-12\right)+8x\)
\(\Rightarrow-6x-31< \left(-12\right)+8x\)
\(\Rightarrow-6x-31+12-8x< 0\)
\(\Rightarrow-6x-8x-31+12< 0\)
\(\Rightarrow\left(-6x-8x\right)-19< 0\)
\(\Rightarrow-14x-19< 0\)
\(\Rightarrow-14x< 0+19\)
\(\Rightarrow-14x< 19\)
\(\Rightarrow x< 19\div\left(-14\right)\)
\(\Rightarrow x< -\frac{19}{14}\)
Vậy \(x< -\frac{19}{14}\)
cảm ơn