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\(2x-49=5.32\\ \Leftrightarrow2x-49=160\\ \Leftrightarrow2x=209\\ \Leftrightarrow x=\dfrac{209}{2}\)
\(200-\left(2x+6\right)=43\\ \Leftrightarrow2x+6=157\\ \Leftrightarrow2x=151\\ \Leftrightarrow x=\dfrac{151}{2}\)
\(135-5\left(x+4\right)=35\\ \Leftrightarrow5\left(x+4\right)=100\\ \Leftrightarrow x+4=20\\ \Leftrightarrow x=16\)
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
a) \(\left|x-2\right|-12=-1\)
\(\Leftrightarrow\left|x-2\right|=11\)
TH1 : \(x-2=11\)
\(\Leftrightarrow x=13\)
TH2 : \(x-2=-11\)
\(\Leftrightarrow x=-9\)
Vậy \(x\in\left\{13;-9\right\}\)
b) \(135-\left|9-x\right|=35\)
\(\Leftrightarrow\left|9-x\right|=100\)
TH1 : \(9-x=100\)
\(\Leftrightarrow x=-91\)
TH2 :\(9-x=-100\)
\(\Leftrightarrow x=109\)
Vậy \(x\in\left\{-81;109\right\}\)
c) \(\left|2x+3\right|=4\)
TH1 : \(2x+3=4\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
TH2 : \(2x+3=-4\)
\(\Leftrightarrow2x=-7\)
\(\Leftrightarrow x=-\frac{7}{2}\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{7}{2}\right\}\)
d) 135(-12 + 247) - 147(135 - 12)= -12.135 + 135.147 - 147.135 + 12.147 = (-12.135 + 12.147) + (135.147 - 147.135) = 12(-135 + 147) + 0 = -12.12 = -144
Bài 1:
a) [3.(-2) - (-8)].(-7) - (-2).(-5) = -7(-6 + 8) - 2.5= -7.2 - 2.5 = -2(7 + 5) = -2.12 = -24
c). 135 - 5( x + 4 ) = 35.
135 - 5 . ( x +4) 35
5 . ( x + 4 ) = 135 - 35 = 100
x + 4 = 100 : 5 =20
x = 20 - 4 = 16
a
2x - 49 = 5.9=45
2x=45+49=94
x=94 : 2
x=47