Tìm x
2^100 . x = 2^103
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\(\frac{2x^2+1}{x+2}=\frac{2x^2+4x-4x-8+9}{x+2}=\frac{2x\left(x+2\right)-4\left(x+2\right)+9}{x+2}=2x-4+\frac{9}{x+2}\)
\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)
Cách 2:
\(\frac{2x^2+1}{x+2}=\frac{2\left(x^2-2^2\right)+9}{x+2}=\frac{2\left(x-2\right)\left(x+2\right)+9}{x+2}=2\left(x-1\right)+\frac{9}{x+2}\)
\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}\)
\(B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)
\(B=\frac{1.2.3...38}{3.4.5...40}.\frac{4.5.6...41}{2.3.4...39}\)
\(B=\frac{2}{39.40}.\frac{40.41}{2.3}\)
\(B=\frac{41}{39.3}=\frac{41}{297}\)
\(cam\in A\)
\(táo\in A\); \(táo\notin B\)
\(ổi\in B\); \(ổi\notin A\)
\(chanh\in B\); \(chanh\notin A\)
\(cam\in B\)
\(CAM\in A\)
\(\text{TÁO}\in A\)
\(\text{ỔI}\in B\)
\(CHANH\in B\)
\(CAM\in B\)
a) 5x + 5x+2 =650
5x ( 1 + 52)=650
5x . 26 = 650
5x = 650 : 26 = 25= 52
=> x = 2
b) (7x + 2)-1 = 3-2
\(\frac{1}{7x+2}=\frac{1}{3^2}\)
\(\Rightarrow7x+2=9\)
\(7x=9-2=7\)
\(x=\frac{7}{7}=1\)
2.(x-3)+3(x-1)=0
=>2x-6+3x-3=0
=>(2x+3x)-6-3=0
=>5x-9=0
=>5x=9
=>x=9/5
Ta có: 2.(x-3)+3.(x-1) = 0
2x-6 + 3x -3 = 0
(2x+3x)-6-3=0
5x - 9 = 0
5x = 9
x = 9 : 5
x = 9/5
2^100.x=2^103
chia cả 2 vế cho 2^100 ta có
x=2^3
x=8
vậy x=8
\(2^{100}.x=2^{103}\)
\(x=2^{103}:2^{100}\)
\(x=2^3\)