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Tính :
1-2+3-4+5-6+...+201-202
=(1-2)+(3-4)+(5-6)+...+(201-202) (có 101 cặp)
=(-1)+(-1)+(-1)+...+(-1)
=(-1).101=-101.
Tìm x là số nguyên :
30-(x+1)=-17
x+1=30-(-17)
x+1=47
x=47-1
x=46
Vậy x=46.
x(|15|-10)=53
x(15-10)=53
x.5=53
x=53:5
x=52=25
Vậy x=25
2x+5=7-x-5
2x+x+5=7-5
3x+5=2
3x=2-5
3x=-3
x=(-3):3
x=-1
Vậy x=-1.
Câu 1:
a) 2(x-3)-3(x-5)=4(3-x)-18
<=> 3x-6-3x+15-12+4x+18=0
<=> 4x+15=0
<=> 4x=-15
<=> x=-15/4
b) -2(2x-8)+3(4-2x)=-57-5(3x-7)
<=> -4x+16+12-6x+57+15x-35=0
<=> -5x+50=0
<=> -5x=-50
<=> x=10
c) 3|2x2-7|=33
<=> |2x2-7|=11
<=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=9\\x^2=-2\end{cases}\Leftrightarrow}x=\pm3}\)
d) có 9x+17=3(3x+2)+11
=> 11 chia hết cho 3x+2
=> 3x+2 thuộc Ư (11)={-11;-1;1;11}
ta có bảng
3x+2 | -11 | -1 | 1 | 11 |
x | -13/3 | -1 | -1/3 | 3 |
Câu 2:
xy-5x+y=17
<=> x(y-5)+(y-5)=12
<=> (y-5)(x+5)=12
=> y-5; x+5 \(\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
lập bảng tương tự câu 1
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
a) 5/17 * 8/-7+8/17*-7/3+-7/3*4/17
-40/119 + 12/17 × -7/3
-40/119 + -28/17 =-236/119
b) -10/13 + 5/17 - 3/13 + 12/17 - 11/20
(5/17+12/17)-(10/13+3/13)-11/20
-11/20
\(x^2+2x+4⋮x+1\)
\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có: \(x^2+2x+4\)
\(=\left(x^2+x\right)+\left(x+1\right)+3\)
\(=x\left(x+1\right)+\left(x+1\right)+3\)
\(=\left(x+1\right)\left(x+1\right)+3\)
Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1
\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)
a) 9x-1=32
( 32 )x-1 = 32
32x-2 = 32
⇒ 2x-2 = 2
2x = 2+2
2x = 4
x = 4 : 2
x = 2
b) 5x+2=625
5x+2= 54
⇒ x+2 = 4
x = 4-2
x = 2
c) 2x: 25= 2
2x:25 = 21
2x = 21 . 25
2x = 26
⇒ x = 6
d) 3x:27=3
3x:33 = 31
3x = 31.33
3x = 34
⇒ x = 4
a) Ta có: \(9^{x-1}=3^2\)
\(\Leftrightarrow3^{2x-2}=3^2\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
b) Ta có: \(5^{x+2}=625\)
\(\Leftrightarrow5^{x+2}=5^4\)
\(\Leftrightarrow x+2=4\)
hay x=2
Vậy: x=2
c) Ta có: \(2^x:2^5=2\)
\(\Leftrightarrow2^{x-5}=2^1\)
\(\Leftrightarrow x-5=1\)
hay x=6
Vậy: x=6
d) Ta có: \(3^x:27=3\)
\(\Leftrightarrow3^x:3^3=3\)
\(\Leftrightarrow3^{x-3}=3^1\)
\(\Leftrightarrow x-3=1\)
hay x=4
Vậy: x=4
2x = 17 + 7 = 24
Vậy, x = 24 : 2 = 12
\(2x-7=17\)
\(2x=17+7\)
\(2x=24\)
\(x=24\div2\)
\(x=12\)
Vậy \(x=12\)