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ta có: ( 2x+1) . (y - 3) = 12
=> 2x + 1 = 12 => 2x = 11 => x = \(\frac{11}{2}\)
y - 3 = 12 => y = 12 + 3 = 15
vậy x = \(\frac{11}{2}\) và y = 15
t i c k nhé!! 865856856878769696896856858788567567479647
(2x+1).(y-3)=12
\(\Rightarrow2x+1;y-3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Mà 2x + 1 lẻ.
Ta có bảng :
| 2x+1 | x | y-3 | y |
| -3 | -2 | -4 | -1 |
| -1 | -1 | -12 | -9 |
| 1 | 0 | 12 | 15 |
| 3 | 1 | 4 | 7 |
32x34x3n=310
3n=310-2-4
3n=34
bài 2:
9<3.x<235
A={4;5;6;...;78}
bài 2 nếu sai thì bảo mk nhé
Bài3:
so sánh 230 và 320
bài 3 chịu ahihi
(2x+1)3=125
=> 2x + 1 = 41,666667
=> 2x = 42,66667
=> x = 21,33333
\(\left(2x+1\right)3=125\)
\(2x+1=125.3\)
\(2x+1=375\)
\(2x=375-1\)
\(2x=374\)
\(x=374:2\)
\(x=187\)
(x+1)(x+2)+(x+3)+...+(x+100)=5750
(x+x+...+x)(1+2+3+...+100)=5750
100x + 5050 =5750
100x =700
x =7
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750
x . 100 + ( 100 + 1 ). 100 : 2 =5750
x .100 + 5050 = 5750
x .100 = 5750 - 5050
x . 100 = 700
x = 700 : 100
x = 7
Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
nên \(x=2002-1=2001\)
Vậy x = 2001
\(A=3+3^2+...+3^{2008}\)
\(\Rightarrow3A=3^2+3^3+...+3^{2009}\)
\(\Rightarrow3A-A=3^{2009}-3\)
\(\Rightarrow2A+3=3^{2009}\)
Vậy n = 2009
a: \(2^{x}\cdot4=128\)
=>\(2^{x}=\frac{128}{4}=32=2^5\)
=>x=5
b: \(x^{15}=x\)
=>\(x^{15}-x=0\)
=>\(x\left(x^{14}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)
c: \(\left(2x+1\right)^3=125\)
=>\(\left(2x+1\right)^3=5^3\)
=>2x+1=5
=>2x=5-1=4
=>\(x=\frac42=2\)
d: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)
=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)
(2x+1)^3=5^3
=>th1: 2x+1=5 =>. 2x=4 =>x=2
th2: 2x+1=-5 => 2x=-6 =>x=-3
vậy x=2 hoặc x=-3
( 2x + 1 )3 = 125
( 2x + 1 )3 = 53
=> 2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2
=))
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=5-1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=4:2\)
\(\Leftrightarrow x=2\)
~ rất vui vì giúp đc bn ~
\(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Study well
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=5-1\)
\(2x=4\)
x = 4 : 2 = 2