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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 12 2023
a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 12 2023
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 1
a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
HD
1
AH
Akai Haruma
Giáo viên
9 tháng 4 2023
Lời giải:
$2^x+2^{x+1}+2^{x+2}+...+2^{x+2020}=2^{2024}-8$
$2^x(1+2+2^2+...+2^{2020})=2^{2024}-8(1)$
$2^x(2+2^2+2^3+...+2^{2021})=2^{2025}-16(2)$
Lấy $(2)$ trừ $(1)$ ta có:
$2^x(2^{2021}-1)=2^{2025}-16-(2^{2024}-8)=2^{2024}(2-1)-8$
$2^x(2^{2021}-1)=2^{2024}-8=2^3(2^{2021}-1)$
$\Rightarrow 2^x=2^3$
$\Rightarrow x=3$
31 tháng 1 2017
a, - 2 .( x + 6 ) + 6 . ( x - 10 ) = 8
- 2x - 12 + 6x - 60 = 8
4x - 72 = 8
4x = 8 + 72
4x = 80
x = 20
b, - 4 . ( 2x + 9 ) - ( - 8x + 3 ) - ( x + 13 ) = 0
- 8x - 36 + 8x - 3 - x - 13 = 0
- x - 52 = 0
x = - 52
c, 7x . ( 2 + x ) - 7x . ( x + 3 ) = 14
7x . ( 2 + x - x - 3 ) = 14
7x . ( - 1 ) = 14
7x = 14 : ( - 1 )
7x = - 14
x = - 2
d, 2 . ( 5 + 3x ) + x = 31
10 + 6x + x = 31
10 + 7x = 31
7x = 31 - 10
7x = 21
x = 3
31 tháng 1 2017
a)-2(x+6)+6(x-10)=8
-2x+-12+6x+-60=8
4x+-72=8
4x=80
x=80:4
x=20
b)-4(2x+9)-(-8x+3)-(x+13)=0
-8x+-36+8x-3+x-13=0
(-8x+8x)+-36+-3+x-13=0
0+-52+x=0
x=0-(-52)
x=-52
c)7x(2+x)-7x(x+3)=14
14x+7x2