Tìm x \(\in\)N biết: (11x+5)\(⋮\)(2x+31)
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a, (2\(x\) -5) + 17 = 6
2\(x\) - 5 = 6 - 17
2\(x\) - 5 = - 11
2\(x\) = - 11 + 5
2\(x\) = -6
\(x\) = -3
b, (-18).\(x\) = - 36
\(x\) = -36 : (-18)
\(x\) = 2
c, 15 - (-11\(x\) - 7) = -22
- 11\(x\) - 7 = 15 + 22
-11\(x\) - 7 = 37
11\(x\) = -37 - 7
11\(x\) = - 44
\(x\) = - 4
a) \(\left(2x-5\right)+17=6\Rightarrow2x-5=-11\Rightarrow2x=-6\Rightarrow x=-3\)
b) \(\left(-18\right)x=-36\Rightarrow x=\left(-36\right):\left(-18\right)=2\)
c) \(15-\left(-11x-7\right)=-22\)
\(\Rightarrow15+11x+7=-22\)
\(\Rightarrow11x=-22-7-15\)
\(\Rightarrow11x=-44\)
\(\Rightarrow x=-44:11=-4\)
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)
\(\left(2x+3\right)^8-\left(2x+3\right)^3=0\)
\(\left(2x+3\right)^3.\text{ }\left[\left(2x+3\right)^5-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x+3\right)^3=0\\\left(2x+3\right)^5-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x+3=0\\\left(2x+3\right)^5=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{3}{2}\\x=-1\end{cases}}}\)
Vậy \(x=-\frac{3}{2}\) hoặc \(x=-1\)
Câu b tương tự
c) \(\left|5-3x\right|=\left|11x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}5-3x=11x+2\\5-3x=-11x-2\end{cases}\Leftrightarrow\orbr{\begin{cases}11x+3x=2-5\\-3x+11x=-2+5\end{cases}\Leftrightarrow\orbr{\begin{cases}14x=-3\\8x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{3}{14}\\x=\frac{3}{8}\end{cases}}}}\)
Vậy \(x=-\frac{3}{14}\)hoặc \(x=\frac{8}{3}\)
a) ( 2 + 1 )x = 45
3x = 45
x = 45 : 3
x = 15
b) ( 2 + 7 )x = 918
9x = 918
x = 918 : 9
x = 102
c) ( 2 + 3 )x = 65
5x = 65
x = 65 : 5
x = 13
d) ( 11 + 22 )x = 66
33 x = 66
x = 66 : 33
x = 2
\(5\left(x+2\right)-x^2-2x=0\)
\(\Rightarrow5\left(x+2\right)-\left(x^2+2x\right)=0\)
\(\Rightarrow5\left(x+2\right)-x\left(x+2\right)=0\)
\(\Rightarrow\left(5-x\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
\(\left(11x+5\right)⋮\left(2x+31\right)\Rightarrow11\left(2x+31\right)-2\left(11x+5\right)⋮\left(2x+31\right)\)
\(\Rightarrow341-10⋮\left(2x+31\right)\Rightarrow331⋮\left(2x+31\right)\Rightarrow2x+31\inƯ\left(331\right)=\left\{1;331\right\}\)
vì \(2x+31\ge31\Rightarrow2x+31=331\Rightarrow x=150\)