tim x thuoc z biet:
a)2x-1+2x+1+2x+2=104
b)3x+2+3x+1+3x+3x-1=120
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Bài 2:
\(A=-2x^2+3x-5\)
\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)
\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)
Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)
a: \(\Leftrightarrow x-3\inƯ\left(21\right)\)
\(\Leftrightarrow x-3\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{0;2;4;6;10;24\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{2;0;18\right\}\)
c: \(\Leftrightarrow2x-1+4⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{1;0\right\}\)
d: \(\Leftrightarrow x^2+x+3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;2\right\}\)
\(a,\)\(3x\left(x+1\right)-2x\left(x+2\right)=1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-4x=1-x\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
\(b,\)\(\frac{1}{3}x^2-4x+2x\left(2-3x\right)=0\)
\(\Leftrightarrow\frac{1}{3}x^2-4x+4x-6x^2=0\)
\(\Leftrightarrow-\frac{17}{3}x^3=0\)
\(\Leftrightarrow x=0\)
a) 17-(2x-11) = 12-3x
2x-11 = 17-12+3x
2x-11 = 5+3x
2x-3x = 5+11
-x = 16
x = -16
b, |2x-3|=7
=>2x-3=7 hoặc 2x-3=-7
=>2x=10 hoặc 2x=-4
=>x=5 hoặc x=-2
c, (2x-1)4 = 16
(2x-1)4 = 24
2x-1=2
2x=3
x=3/2
d, (17-2x)3 = -125
(17-2x)3 = (-5)3
17-2x=-5
2x=17-(-5)
2x=22
x=11
a, => 17-2x+11 = 12-3x
=> 28-2x=12-3x
=> 28=12-3x+2x = 12-x
=> x=12-28 = -16
b, => 2x-3=7 hoặc 2x-3=-7
=> x=5 hoặc x=-2
c, => 2x-1=2 hoặc 2x-1=-2
=> x=3/2 hoặc x=-1.2
d, => 17-2x=-5
=> 2x=17-(-5) = 22
=> x=22:2 = 11
Tk mk nha
a) \(2^{x-1}+2^{x+1}+2^{x+2}=104\)
=> \(2^{x-1}+2^x\cdot2+2^x\cdot2^2=104\)
=> \(2^x:2+2^x\cdot\left(2+2^2\right)=104\)
=> \(2^x\cdot\frac{1}{2}+2^x\cdot6=104\)
=> \(2^x\cdot\left(\frac{1}{2}+6\right)=104\Rightarrow2^x=104:\left(\frac{1}{2}+6\right)=104:\frac{13}{2}=16\)
=> \(x=4\)
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