3^n+1-2x3^n + 2^n+5-7x2^n⋮25
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ZP
1
HP
5 tháng 4 2016
3n+3-2.3n+2n+5-7.2n
=3n.33-2.3n+2n.32-7.2n
=3n.(33-2)+2n.(32-7)
=3n.25+2n.25=25.(3n+2n) luôn chia hết cho 25
NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 6
a: \(2x^3-50x=0\)
=>\(2x\left(x^2-25\right)=0\)
=>x(x-5)(x+5)=0
=>x∈{0;5;-5}
b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
=>2x(3x-5)+(3x-5)=0
=>(3x-5)(2x+1)=0
=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)
c: \(9\left(3x-2\right)=x\left(2-3x\right)\)
=>9(3x-2)-x(2-3x)=0
=>9(3x-2)+x(3x-2)=0
=>(3x-2)(x+9)=0
=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)
d: \(\left(2x-1\right)^2-25=0\)
=>(2x-1-5)(2x-1+5)=0
=>(2x-6)(2x+4)=0
=>(x-3)(x+2)=0
=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)
e: \(25x^2-2=0\)
=>\(25x^2=2\)
=>\(x^2=\frac{2}{25}\)
=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)
f: \(x^2-25=6x-9\)
=>\(x^2-6x-16=0\)
=>(x-8)(x+2)=0
=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)
g: 5x(x-3)-2x+6=0
=>5x(x-3)-2(x-3)=0
=>(x-3)(5x-2)=0
=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)
h: 3x(x-7)-2(x-7)=0
=>(x-7)(3x-2)=0
=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)
i: \(7x^2-28=0\)
=>\(7x^2=28\)
=>\(x^2=4\)
=>x=2 hoặc x=-2
j: 2x+1+x(2x+1)=0
=>(2x+1)(x+1)=0
=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)
k: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
=>(x+2)(x+2-x+2)=0
=>4(x+2)=0
=>x+2=0
=>x=-2
l: \(x^3+5x^2-4x-20=0\)
=>\(x^2\left(x+5\right)-4\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^2-4\right)=0\)
=>(x+5)(x-2)(x+2)=0
=>x∈{-5;2;-2}
m: \(x^2-25+2\left(x+5\right)=0\)
=>(x-5)(x+5)+2(x+5)=0
=>(x+5)(x-3)=0
=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)
n: \(x^2-3x+2=0\)
=>\(x^2-x-2x+2=0\)
=>x(x-1)-2(x-1)=0
=>(x-1)(x-2)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)
o: \(x^2-6x+8=0\)
=>\(\left(x-2\right)\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)
p: \(x^2-5x-14=0\)
=>\(x^2-7x+2x-14=0\)
=>(x-7)(x+2)=0
=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)
q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
=>\(x^2-4x+4-x^2+9=6\)
=>-4x+13=6
=>-4x=6-13=-7
=>x=7/4
r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)
=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)
=>-4x+26=18
NL
Nguyễn Lê Phước Thịnh
CTVHS
20 tháng 7 2021
a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
8 tháng 1 2017
3^2.3^n=3^5
=> n=3
(2^2:4).n^2=4
=> n= +2;-2
1/9.3^4.3^n=3^7
=>n=9
1/9=27^n=3^n
=> n=
1/2.2n+4.2^n=9.5^n
=> n ∈ ∅
NL
Nguyễn Lê Phước Thịnh
CTVHS
14 tháng 11 2025
a: Sửa đề: \(A\left(x\right)=-4x^2-2x-8+5x^3-7x^2+1\)
Ta có: \(A\left(x\right)=-4x^2-2x-8+5x^3-7x^2+1\)
\(=5x^3+\left(-4x^2-7x^2\right)-2x-8+1=5x^3-11x^2-2x-7\)
\(B\left(x\right)=-3x^3+11x^2+9+x-2x-2x^3\)
\(=-3x^3-2x^3+11x^2+x-2x+9=-5x^3+11x^2-x+9\)
b: N(x)=A(X)+B(x)
\(=5x^3-11x^2-2x-7-5x^3+11x^2-x+9=-3x+2\)
c: N(x)=0
=>-3x+2=0
=>-3x=-2
=>\(x=\frac23\)
8 tháng 1 2017
a)\(3^2 . 3^n = 3^5\)
=> \( n=3\)
b) \((2^2:4) . n^2=4\)
=> \(n=-2;2\)
c)\(\frac{1}{9}.27^n=3^n\)
=> n ∈ Z
d) \(\frac1{9}.3^4.3^n=3^7\)
=> n=5
e)n ∈ ∅
\(3^{n+1}-2.3^n+2^{n+5}-7.2^n\)
\(=3^n\left(3-2\right)+2^n\left(2^5-7\right)\)
\(=3^n+2^n.25\)
Vì \(3^n⋮̸25\); \(25.2^n⋮25\)
=> \(3^n+2^n.25⋮̸25\)
=> \(3^{n+1}-2.3^n+2^{n+5}-7.2^n⋮̸25\)