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\(2^{3x}:2^{x+3}=32\Rightarrow2^{3x-x-3}=2^5\Rightarrow x^{2x-3}=2^5\Rightarrow2x-3=5=>2x=8=>x=4\)
\(2^{5x}:2^{4x}=32\Rightarrow2^{5x-4x}=2^5=>2^x=2^5=>x=5\)
\(\left(x-\frac{1}{2}\right)^3=8=2^3\Rightarrow x-\frac{1}{2}=2\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}\)
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x-32+15-3x=-4x+1-32
x-3x+4x=1-32+32-15
2x=1+32-32-15
2x=1+(32-32)-15
2x=1-15
2x=-14
x=-14:2
x= - 7
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a, 3x - 5 = 16...............
=> 3x = 16 + 5 = 21..................
=> x = 21 : 3 = 7..........
\(b,42-2\left(32-2^{x+1}\right)=10\)
\(\Rightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Rightarrow32-2^{x+1}=32:2=16\)
\(\Rightarrow2^{x+1}=32-16=16=2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1=3\)
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a, 3x - 2 ⋮ x + 3
=> 3x + 9 - 11 ⋮ x + 3
=> 3(x + 3) - 11 ⋮ x + 3
=> 11 ⋮ x + 3
b, x ⋮ 2x + 1
=> 2x ⋮ 2x + 1
=> 2x + 1 - 1 ⋮ 2x + 1
=> 1 ⋮ 2x + 1
c, 3x + 6 ⋮ x + 1
=> 3x + 3 + 3 ⋮ x + 1
=> 3(x + 1) + 3 ⋮ x + 1
=> 3 ⋮ x + 1
d, em không biết làm
câu a,b,c bn Cả Út lm r
mik làm câu d
\(x^2⋮x-2\)
\(\Rightarrow x\left(x-2\right)+2x⋮x-2\)
\(\Rightarrow2x⋮x-2\)
\(\Rightarrow2\left(x-2\right)+4⋮x-2\)
\(\Rightarrow4⋮x-2\)
\(\Rightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{3;1;4;0;6;-2\right\}\)
Vậy..............................
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\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)
\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)
\(\Leftrightarrow2,8x=-60+32=-28\)
\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)
d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)
Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.
mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha
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\(\left(3x+1\right)^8>=0\)
\(\left(2y^2-32\right)^8>=0\)
Do đó: \(\left(3x+1\right)^8+\left(2y^2-32\right)^8>=0\)
Dấu '=' xảy ra khi 3x+1=0 và 2y2-32=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y\in\left\{4;-4\right\}\end{matrix}\right.\)
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\(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)
\(\Rightarrow-7x=\frac{-1}{6}\)
\(\Rightarrow x=\frac{1}{42}\)
Vậy ...
\(\)
\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{18}\)
Vậy...
\(2^{3x+1}=32\)
\(2^{3x+1}=2^5\)
TỪ PHÉP TÍNH TRÊN, SUY RA :
\(3x+1=5\)
\(3x=5-1\)
\(3x=4\)
\(x=4:3\)
\(x=\frac{4}{3}\)
VẬY \(x=\frac{4}{3}\)
\(2^{3x+1}=32\)
\(\Rightarrow2^{3x+1}=2^{5^{ }}\)
\(\Rightarrow3x+1=5\)
\(3x=5-1\)
\(3x=4\)
\(x=4\div3\)
\(x=\frac{4}{3}\)