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c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
13/ => 10 + 2x = 42 = 16
=> 2x = 6
=> x = 3
14/ => 52x : 53 - 50 = 75
=> 52x : 53 = 125 = 53
=> 52x = 56
=> 2x = 6
=> x = 3
15/ => (26 - 3x) : 5 = 4
=> 26 - 3x = 20
=> 3x = 6
=> x = 2
16/ => x - 17 = -25
=> x = -8
Bài 1 :
\(2^x.8=512\)
\(2^x=512:8\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
\(b,\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(c,x^{20}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(d,\left(x-3\right)^{10}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) Để \(-1:x\)là số nguyên
\(\Rightarrow\)\(x\inƯ\left(-1\right)\in\left\{\pm1\right\}\)
Vậy \(x\in\left\{-1;1\right\}\)
b) Để \(1:x+1\)là số nguyên
\(\Rightarrow\)\(x+1\inƯ\left(1\right)\in\left\{\pm1\right\}\)
+ \(x+1=1\)\(\Leftrightarrow\)\(x=1-1=0 \left(TM\right)\)
+ \(x+1=-1\)\(\Leftrightarrow\)\(x=-1-1=-2\left(TM\right)\)
Vậy \(x\in\left\{-2; 0\right\}\)
c) Để \(-2:x\)là số nguyên
\(\Rightarrow\)\(x\inƯ\left(-2\right)\in\left\{\pm1;\pm2\right\}\)
Vậy \(x\in\left\{-1;-2;1;2\right\}\)
d) Để \(3:x-2\)là số nguyên
\(\Rightarrow\)\(x-2\inƯ\left(3\right)\in\left\{\pm1;\pm3\right\}\)
- Ta có bảng giá trị:
| \(x-2\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
| \(x\) | \(1\) | \(3\) | \(-1\) | \(5\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-1;1;3;5\right\}\)
e) Ta có: \(x+8=\left(x-7\right)+15\)
- Để \(x+8⋮x-7\)\(\Leftrightarrow\)\(\left(x-7\right)+15⋮x-7\)mà \(x-7⋮x-7\)
\(\Rightarrow\)\(15⋮x-7\)\(\Rightarrow\)\(x-7\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
- Ta có bảng giá trị:
| \(x-7\) | \(-1\) | \(1\) | \(-3\) | \(3\) | \(-5\) | \(5\) | \(-15\) | \(15\) |
| \(x\) | \(6\) | \(8\) | \(4\) | \(10\) | \(2\) | \(12\) | \(-8\) | \(22\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-8;2;4;6;8;10;12;22\right\}\)
f) Ta có: \(2x+9=\left(2x-10\right)+19=2.\left(x-5\right)+19\)
- Để \(2x+9⋮x-5\)\(\Leftrightarrow\)\(2.\left(x-5\right)+19⋮x-5\)mà \(2.\left(x-5\right)⋮x-5\)
\(\Rightarrow\)\(19⋮x-5\)\(\Rightarrow\)\(x-5\inƯ\left(19\right)\in\left\{\pm1;\pm19\right\}\)
- Ta có bảng giá trị:
| \(x-5\) | \(-1\) | \(1\) | \(-19\) | \(19\) |
| \(x\) | \(4\) | \(6\) | \(-14\) | \(24\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-14;4;6;24\right\}\)
g) Ta có: \(2x+16=\left(2x-16\right)+32=2.\left(x-8\right)+32\)
- Để \(2x+16⋮x-8\)\(\Leftrightarrow\)\(2.\left(x-8\right)+32⋮x-8\)mà \(2.\left(x-8\right)⋮x-8\)
\(\Rightarrow\)\(32⋮x-8\)\(\Rightarrow\)\(x-8\inƯ\left(32\right)\in\left\{\pm1;\pm2;\pm4;\pm8;\pm16;\pm32\right\}\)
- Ta có bảng giá trị:
| \(x-8\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-4\) | \(4\) | \(-8\) | \(8\) | \(-16\) | \(16\) | \(-32\) | \(32\) |
| \(x\) | \(7\) | \(9\) | \(6\) | \(10\) | \(4\) | \(12\) | \(0\) | \(16\) | \(-8\) | \(24\) | \(-24\) | \(40\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-24;-8;0;4;6;7;9;10;12;16;24;40\right\}\)
h) Ta có: \(5x+2=\left(5x-5\right)+7=5.\left(x-1\right)+7\)
- Để \(5x+2⋮x-1\)\(\Leftrightarrow\)\(5.\left(x-1\right)+7⋮x-1\)mà \(5.\left(x-1\right)⋮x-1\)
\(\Rightarrow\)\(7⋮x-1\)\(\Rightarrow\)\(x-1\inƯ\left(7\right)\in\left\{\pm1;\pm7\right\}\)
- Ta có bảng giá trị:
| \(x-1\) | \(-1\) | \(1\) | \(-7\) | \(7\) |
| \(x\) | \(0\) | \(2\) | \(-6\) | \(8\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-6;0;2;8\right\}\)
k) Ta có: \(3x=\left(3x-6\right)+6=3.\left(x-2\right)+6\)
- Để \(3x⋮x-2\)\(\Leftrightarrow\)\(3.\left(x-2\right)+6⋮x-2\)mà \(3.\left(x-2\right)⋮x-2\)
\(\Rightarrow\)\(6⋮x-2\)\(\Rightarrow\)\(x-2\inƯ\left(6\right)\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
- Ta có bảng giá trị:
| \(x-2\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-3\) | \(3\) | \(-6\) | \(6\) |
| \(x\) | \(1\) | \(3\) | \(0\) | \(4\) | \(-1\) | \(5\) | \(-4\) | \(8\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-4;-1;0;1;3;4;5;8\right\}\)
bài 1 :
a) S1=( 1 + 3 - 5 - 7 )+(9+11-13-15)+...+(393+395-397-399)
S1=(-8)+(-8)+...+(-8)
S1=(-8)*199
S1=-1592
b)S2=(1-2-3+4)+( 5 - 6 - 7 +8)+...+( 97 - 98 - 99 + 100)
S2=0+0+...+0
S2=0*100
S2=0
phần c và d tương tự nhé
BÀI 2
c)<=>2(x-1)+4 chia hết x-3
=>8 chia hết x-3
=>x-3\(\in\){-1,-2,-4,-8,1,2,4,8}
=>x\(\in\){2,1,-1,-5,4,5,7,11}
a,\(15-\left(-4-x\right)=6\)
\(\Leftrightarrow-4-x=15-6\)
\(\Leftrightarrow-4-x=9\)
\(\Leftrightarrow x=-4-9\)
\(\Leftrightarrow x=-13\)
b,\(-30+\left(25-x\right)=-1\)
\(\Leftrightarrow25-x=-1-\left(-30\right)\)
\(\Leftrightarrow25-x=29\)
\(\Leftrightarrow x=25-29=-4\)
c,\(x-\left(12-25\right)=-8\)
\(\Leftrightarrow x+13=-8\)
\(\Leftrightarrow x=-21\)
e,\(x-5=1\Rightarrow x=6\)
g,\(x+30=-4\)
\(\Leftrightarrow x=-4-30\)
\(\Leftrightarrow x=-34\)
h,\(x-\left(-24\right)=3\)
\(\Leftrightarrow x+24=3\)
\(\Leftrightarrow x=3-24\)
\(\Leftrightarrow x=-21\)
i,\(\left(x+5\right)+\left(x-9\right)=x+2\)
\(\Leftrightarrow x+5+x-9-x-2=0\)
\(\Leftrightarrow x-6=0\)
\(\Leftrightarrow x=6\)
k,\(\left(27-x\right)+\left(15+x\right)=x-24\)
\(\Leftrightarrow27-x+15+x-x+24=0\)
\(\Leftrightarrow-x+66=0\)
\(\Leftrightarrow-x=-66\)
\(\Leftrightarrow x=66\)
a) 15-(-4-x)=6
15 + 4 + x = 6
19 + x = 6
x = 6 - 19
x = 6 + ( -19 )
x = -13
b) -30+( 25- x ) =-1
25- x = -1 - ( - 30 )
25- x = -1 + 30
25- x =29
x = 25 - 29
x = 25 + ( -29 )
x = -4
c) x- ( 12-25) =-8
x - ( - 13 ) = -8
x + 13 = -8
x = -8 - 13
x = -8 + ( -13 )
x = - 21
d) ( x-29 ) - (17- 38 ) = -9
( x-29) - (-21 ) = -9
x -29 = -9 + ( -21 )
x -29 = -30
x = -30 + 29
x = -1
e) x-5=1
x = 1 + 5
x = 6
g) x + 30 = -4
x = -4 - 30
x = -4 + ( -30 )
x = -34
h) x- ( -24 ) = 3
x = 3 + ( -24 )
x = -21
l) 22- ( - x ) = 12
-x = 22 - 12
-x = 10
x = -10
i) ( x + 5 ) + ( x - 9 ) = x + 2
x + 5 + x - 9 = x + 2
x + x + ( 5 - 9 ) = x + 2
x + x + ( -4 ) = x + 2
x + x = x + 2 - ( -4 )
x + x = x + 2 + 4
x + x =x + 6
=> x = 6
1. Tìm số nguyên x, biết :
a) x : 13 = -3
x=-3.13
x=-39
b) 2x - ( -17 ) = 15
2x+17=15
2x=15-17
2x=-2
x=-2:2
x=-1
c) x - 42 = -3
x=-3+42
x=39
d) 3x - 27 = 9
3x=9+27
3x=36
x=36:3
x=12
e) 2x + 12 = 3( x - 17 )
2x+12=3x-51
12+51=3x-2x
73=x
g) 2x2 - 1 = 49
2x2=49+1
2x2=50
x2=50:2
x2=25
x2=52
=> x=5 hoặc x=-5
h) | x + 9 | . 2 = 10
|x+9|=10:2
|x+9|=5
* x+9=5 * x+9=-5
x=5-9 x=-5-9
x=-4 x=-14
a) x : 13 = -3
=> x = -3 . 13 = - 39
Vậy x = - 39
b) 2x - ( -17 ) = 15
=> 2x + 17 = 15
=> 2x = - 2
=> x = -1
Vậy x = -1
c) x - 42 = -3
=> x = -3 + 42 = 39
Vậy x = 39
d) 3x - 27 = 9
=> 3x = 9 + 27 = 36
=> x = 12
Vậy x = 12
e) 2x + 12 = 3( x - 17 )
=> 2x + 12 = 3x - 51
=> 12 + 51 = 3x - 2x
=> x = 63
Vậy x = 63
g) 2x2 - 1 = 49
=> 2x2 = 50
=> x2 = 25
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
h) | x + 9 | . 2 = 10
=> |x + 9 | = 5
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy ....
Dài quá
@@ Học tốt
## Chiyuki Fujito