4x-5 chia het x+1
Tìm xeZ
a) (2x + 5) chia hết (x+2)
b) (3x+5) chia het (x-2)
c) (2-4x) chia het ( x-1)
d) ( x^2 - x+2) chia het (x-1)
a: \(\Leftrightarrow2x+4+1⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1\right\}\)
hay \(x\in\left\{-1;-3\right\}\)
b: \(\Leftrightarrow3x-6+11⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{3;1;13;-9\right\}\)
c: \(\Leftrightarrow-4x+4-2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{2;0;3;-1\right\}\)
d: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{2;0;3;-1\right\}\)
Tìm xeZ
a) (2x + 5) chia hết (x+2)
b) (3x+5) chia het (x-2)
c) (2-4x) chia het ( x-1)
d) ( x^2 - x+2) chia het (x-1)
MK lm mẫu cho câu a) nhé, các câu còn lại bn làm tương tự
a) \(2x+5\)\(⋮\)\(x+2\)
\(\Rightarrow\)\(2\left(x+2\right)+1\)\(⋮\)\(x+2\)
Ta thấy \(2\left(x+2\right)\)\(⋮\)\(x+2\)
nên \(1\)\(⋮\)\(x+2\)
\(\Rightarrow\)\(x+2\)\(\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow\)\(x=\left\{-3;-1\right\}\)
Vậy...
a) \(2x+5\)\(⋮\)\(x+2\)
\(\Rightarrow\)\(2\left(x+2\right)+1\)\(⋮\)\(x+2\)
Ta thấy \(2\left(x+2\right)\)\(⋮\)\(x+2\)
nên \(1\)\(⋮\)\(x+2\)
\(\Rightarrow\)\(x+2\)\(\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow\)\(x=\left\{-3;-1\right\}\)
(x-3) chia het cho (x+1) ; (2x+5) chia het cho (x+1); (4x+1)chia het cho (2x+2)
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tim x
4x+ 5 chia het 2x - 1
4x + 5 ⋮ 2x - 1
<=> 4x - 2 + 7 ⋮ 2x - 1
<=> 2(2x - 1) + 7 ⋮ 2x - 1
=> 7 ⋮ 2x - 1 hay 2x + 1 thuộc ước của 7
Ư(7) = { - 7 ; - 1 ; 1 ; 7 }
Ta có bảng sau :
2x - 1 | - 7 | - 1 | 1 | 7 |
x | - 3 | 0 | 1 | 4 |
Vậy x = { - 3 ; 0 ; 1 ; 4 }
4x + 5 \(⋮\)2x - 1
=> (4x - 2) + 7 \(⋮\)2x - 1
=> 2(2x - 1) + 7 \(⋮\)2x - 1
=> 7 \(⋮\)2x - 1
=> 2x - 1 \(\in\)Ư(7) = {-7;-1;1;7}
=> 2x \(\in\){-6;0;2;8}
=> x \(\in\){-3;0;1;4}
Tìm cac s số nguyên x thỏa mãn
(x+4) chia het (x+1)
(4x+3) chia het (x-2)
2x + 5 chia hết cho 2x - 1
2x chia het cho x-3
2x-1chia hết cho 3x
4x + 7 chia hết cho 3x-5
Ta có : 2x + 5 chia hết cho 2x - 1
=> 2x - 1 + 6 chia hết cho 2x - 1
=> 6 chia hết cho 2x - 1
=> 2x - 1 thuộc Ư(6) = {1;2;3;6}
=> 2x thuộc {2;4}
=> x = {1;2}
Vậy x = {1;2}
a)Ta có : 2x + 5 \(⋮\) cho 2x - 1
=> 2x - 1 + 6 \(⋮\)cho 2x - 1
=> 6 \(⋮\) cho 2x - 1
=> 2x - 1 \(\in\) Ư(6) = {1;2;3;6}
=> 2x \(\in\){2;4}
=> x = {1;2}
Vậy x = {1;2}
Ta có : 2x + 5 chia hết cho 2x - 1
=> 2x - 1 + 6 chia hết cho 2x - 1
=> 6 chia hết cho 2x - 1
=> 2x - 1 thuộc Ư(6) = {1;2;3;6}
=> 2x thuộc {2;4}
=> x = {1;2}
Vậy x = {1;2}
1. tim x
a) x+4 chia het x+1 b) x-7 chia het x-3 c) (4x+3) chia het x+2 d) 4x-5 chia het x
2. tim x,y
a) (x-3).(2y+1)=7
b) (2x+1).(3y-2)=-55
3. tim x
a) (x+1)+(x+3)+(x+5)+...+(x+99)=0
b) (x-3)+(x-2)+(x-1)+...+10+11=11
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(4x +5) chia het (2x-8)
tim x
4x+5 chia hết 2x-8
4x-16 chia hết 2x-8
=>21 chia hết 2x-8
=>2x-8 thuộc {21,7,3,1,-1,-3,-7,-21}
Xét các TH, ta có x=14,5 : 7,5 :5,5 ;4,5;3,5;0,5;-6,5
Tìm x thuộc Z
a/ 7 chia hết x-4
b/ x+8 là Ư(-6)
c/ (-11) chia het x-1
d/ 4x-5 chia hết x
e/ 2x-1 là Ư(3x+2)
f/ x-1 chia hết x+5
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