\(3^{4x-1}:3^x=243\)
Tìm x,y biết:
a) 3 4 x + 1 5 = 1 6 b) 1 3 + 1 2 : x = − 4 c) c ) x − 2 3 x + 1 4 = 0
d) 2 x − 3 3 4 x + 1 = 0 d) 2 3 x + 1 2 − 5 3 = 0 e) 2 x − 0 , 3 = 1 , 5
f) x . − 3 5 3 = − 3 5 5 g) x : 1 3 2 = 1 3 3 h) x 26 = 21 39
k) 1 3 x = 2 3 1 6 l) 77 7 9 5 9 = x 0 , 03 m) 2 2 3 0 , 03 = 1 7 9 x
n) 2 x − 3 2 = 16 o) 3 x − 2 5 = − 243
tim x biết
3x+4=0
2x*(x-1)-(1+2x)=-34
X^2+9x-10=0
(7x-1)*(2+5x)=0
\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
x^4 - 4x^3 + 3x^2 - 34x - 7 =0
3x^4 +2x^3 -34x^2 +2x +3 =0
tìm x
a) 5x(x-3)(x+3)-(2x-3)^2- 5(x+2)^2 +34x(x+2)=1
b) (x-2)^3+ 6(x+1)^2 - (x-3)(x^2+3x+9)=97
tìm x
a) 5x(x-3)(x+3)-(2x-3)^2- 5(x+2)^2 +34x(x+2)=1
b) (x-2)^3+ 6(x+1)^2 - (x-3)(x^2+3x+9)=97
3x*3 mu x+1 =243
3x*3x+1=243
=>3x+2=243
=>3x+2=35
=>x+2=5
=>x=3
Phân tích đa thức thành nhân tử:
a) 12x3+8x2-3x-2
b) 18x3+27x2-2x-3
c) 8x3+4x2-34x+15
e) x3+15x2-34x-12
a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
34x+2 + 34x+2 + 34x+2
Tìm X ∈ N biết:
a,2x . 2x + 1=128
b,3x+1 . 3x=243
a)2x.2x+1=128
\(4x^2+1=128\)
\(\Rightarrow4x^2=127\)
\(\Rightarrow x^2=127\div4\)
\(\Rightarrow x^2=3.175\)
mà \(x\in N\) hay\(x^2\in N\)
\(\Rightarrow\)Không có x
b) 3x+1.3x=243
\(\Rightarrow\)3x+3x=243
\(\Rightarrow\)6x=243
\(\Rightarrow\)x=40,5
Mà \(x\in N\)
\(\Rightarrow\)Không có x