(4x-8)(12-y) =0
(2x-12)-14=86
(127-x) -17=43
Tìm số tự nhiên xbiết:
a) (x – 45).27 = 0
b) 41.(2x – 8).13 = 0
c) (x – 36):18 = 12
d) 43.(x – 18) = 86
e) 96 : (121 – x) = 4
f) 340 : (15x+5) = 17
a) (x – 45).27 = 0 ó x – 45 = 0 ó x = 45
b) 41.(2x – 8).13 = 0 ó 2x – 8 = 0 ó x = 4
c) (x – 36):18 = 12 ó x – 36 = 12.18 ó x = 216 + 36 ó x = 252
d) 43.(x – 18) = 86 ó x – 18 = 86:43 ó x = 2+18 ó x = 20
e) 96 : (121 – x) = 4 ó 121 – x = 96 : 4 ó 121 – x = 24 ó x = 121 – 24 ó x = 97
f) 340 : (15x+5) = 17 ó 15x + 15 = 340 :17 ó 15x + 5 = 20 ó 15x = 20 – 5 ó 15x = 15 ó x = 1
Tìm x sao cho x thuộc tập hợp số nguyên:
1) x - 43 = (35 - x) - 48
2) 305 - x + 14 = 48 + (x + 23)
3) - (x - 6 + 85) = (x + 51) - 54
4) - (35 - x - 37 - x) = 33 - x
5) 13 - | x | = | -4 |
6) | x | - 3 + 6 = 16
7) 35 - | 2x - 1 | = 14
8) | 3x - 2 | + 5 = 9 - x
9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )
10) | 17 + ( x - 15 ) | < 4
11) x2 - 5x = 0
12) | x-9 | . (-8) = -16
13) | 4 - 5x = 24 với x < hoặc = 0
14) x . ( x - 2 ) > 0
15) x . ( x - 2 ) < 0
16) (x-1) . (y+1) = 5
17) x . ( y +2 ) = -8
18) xy - 2x - 2y = 0
19) 2x - 5 chia hết cho x - 1
1) x - 43 = (35 - x) - 48
=> x + x = 35 - 48 + 43
=> x + x = 30
=> x = 30 : 2
=> x = 15
2) 305 - x + 14 = 48 + (x + 23)
=> 305 - x + 14 = 48 + x + 23
=> -x - x = 48 + 23 - 14 - 305
=> -x - x = -248
=> -x = -248 : 2
=> -x = -124
=> x = 124
3) - (x - 6 + 85) = (x + 51) - 54
=> -x + 6 - 85 = x + 51 - 54
=> -x - x = 51 - 54 + 85 - 6
=> -x - x = 76
=> -x = 76 : 2
=> -x = 38
=> x = -38
4) - (35 - x - 37 - x) = 33 - x
=> -35 + x + 37 + x = 33 - x
=> x + x + x = 33 + 35 - 37
=> x + x + x = 31
=> x = 31 : 3
=> x \(=\dfrac{31}{3}\)
Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.
5) 13 - | x | = | -4 |
=> 13 - |x| = 4
=> |x| = 13 - 4
=> |x| = 9
=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
6) | x | - 3 + 6 = 16
=> |x| = 16 - 6 + 3
=> |x| = 13
=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)
7) 35 - | 2x - 1 | = 14
=> |2x - 1| = 35 - 14
=> |2x - 1| = 21
=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)
8) | 3x - 2 | + 5 = 9 - x
=> |3x - 2| = 9 - 5 - x
=> |3x - 2| = 4 - x
=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)
Vì x \(\in\) Z nên x = -1.
9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )
=> x - (-18) > 12 - 1
=> x + 18 > 11
=> x > 11 - 18
=> x > -7
10) | 17 + ( x - 15 ) | < 4
=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)
11) x2 - 5x = 0
=> x . (2 - 5) = 0
=> x . (-3) = 0
=> x = 0 : (-3)
=> x = 0
12) | x-9 | . (-8) = -16
=> |x - 9| = (-16) : (-8)
=> |x - 9| = 3
=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)
13) | 4 - 5x | = 24 với x < hoặc = 0
=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vì x \(\le\) 0 nên x = -4
14) x . ( x - 2 ) > 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x>2\\x< 2\end{matrix}\right.\)
15) x . ( x - 2 ) < 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)
16) (x-1) . (y+1) = 5
=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
17) x . ( y +2 ) = -8
=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=2\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-2\\y+2=4\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=4\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-4\\y+2=2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=2-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
18) xy - 2x - 2y = 0
=> x . (y - 2) - 2y = 0
=> x . (y - 2) - 2y - 4 = -4
=> x . (y - 2) - 2 . (y - 2) = -4
=> (y - 2) . (x - 2) = -4
=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=2\\x-2=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2+2\\x=-2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=4\\x=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-2\\x-2=2\end{matrix}\right.=>\left[{}\begin{matrix}y=-2+2\\x=2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)
19) 2x - 5 \(⋮\) x - 1
=> (2x - 2) - (5 - 2) \(⋮\) x - 1
=> 2(x - 1) - 3 \(⋮\) x - 1
Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1
=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}
=> x \(\in\) {-2; 0; 2; 4}
P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~
B) 17X - ( -16X - 37 ) + 2x + 43
C) -2x -3.( x-17) = 34 -2 (-2x +25)
D) 17x+ 3. ( -16x -37) = 2x +43 -4x
E) {- 3x +2. [45 - x-3 (3x + 7) - 2x] + 4x} =55
F) -103 - 57 : [ -2 ( 2x -1 )2 - ( -9 )0 ] - 106
G) -2x + 3. { 12-2 [ 3x - ( 20 + 2x ) - 4x ] + 1} = 45
Thu gọn biểu thức
a) M= x 2 + 4 x − 1 − 2 x + 8 − 1 x + 4 khi x ≥ 1 4 ;
b) N = − 8 x 3 + 12 x 2 + 1 + 2 x + 4 x khi − 1 2 ≤ x ≤ 0 .
Tìm số tự nhiên x biết:
a) x - 45 . 27 = 0 b) 41 . 2 x - 8 . 13 = 0 c) x - 36 : 18 = 12
d) 43 . x - 18 = 86 e) 96 : 121 - x = 4 f) 340 : 15 x + 5 = 17
5. 91/27 - 56/9 : x = 154/7
4. 1 - ( 43/8 + x - 173/24 ) : 50/3 = 0
-8/5 : 4/5 + | x | = 0,125 * ( -4 )| x + 1/3 | * 17/5 - 12/5 = 21/2x - 2/3x = 7/12 + 3/4x
1.thực hiện phép tính bằng cách hợp lí
a.(-46)+(-125)+46+25 b.25.(-15)+25.(-5)+(-20).75
c.(-151)+(-37)+(-42)+(-63)+142 d.32+(-149)+(-311)+(-89)+(-51)
e.-65.(87-17)-87.(17-65) g.-43.(53-16) - 53.(16-43)
2.tím x thuộc z biết
a.(-5)^2-(5x-3)=4 d.3x^2=12
b.5x-15=-75-x e.|2x-3|=7
c.-2x-40=(5-x)-(-15+60) g. 19-(43-|x|)=45
h.|x-1|+(-5)=2 i.-10-|5-x|=-12
f.32-8.|2x-9|=24 k.4.|x-5|-54=-34 l.259-51.|4x+3|=(-10)^2.3+10^1
3.tìm các số nguyên x,y biết
a.xy+3x-y-5=0
b.xy-2x+y-3=0
1. Thực hiện phép tính bằng cách hợp lí :
a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]
= 0+(-100) = -100
b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75
= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000
c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]
= (-151) + [(-100) + 100] = -151
d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]
= 32+[(-200)+(-400)] = 32+(-600) = -568
e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65
= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)
= 65.(-70) = -4550
g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43
= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)
= 16.(-10) = -160
1. Thực hiện phép tính (tính hợp lý nếu có thể):
a) (-32).43+(-32).65-(-32).8
b) (-43).25+25.(-19)+25.(-38)
c) (-26).13+86.(-26)+(-26)
d) (-17).39+(-17)+(-17).60
2. Tìm số nguyên x, biết:
a) x+5+2.x=17
b) x-11=2.x+4
c) (3-x).(x+5)=0
d) (2.x+2).(x-19)=0
e) (x+2)3=(-125)
f) |x-3|=4
g) 2.|7-x|=16
h) 12-2.|x-10|=(-18)
3. Tìm số nguyên x, biết:
a) 5/8=x/16
b) x/6=1/(-3)
c) x+1/3=20/(-12)
d) 4/5=(-12)/9-x
e) x/2=8/x
f) -x/3=(-12)/x
g) 5/7= -2x/14
h) 5-x/2=2/5-x
4. Tìm các số nguyên x, y, biết:
a) (x-2).(y+1)=5
b) (3-x).(2.y+5)=4
c) x-x.y-y=2
5. Tìm số nguyên n, biết:
a) n+5 ⋮ n-1
b) 2.n-3 ⋮ n+4
c) 3.n+4 ⋮ 2.n-1
ban chia ra tung bai di dai lam
bai nao lam dc thi giam di nhe
Bài 4:
a) \(\left(x-2\right)\left(y+1\right)=5\)
\(\Leftrightarrow x-2;y+1\inƯ\left(5\right)\)
\(\Leftrightarrow x-2;y+1\in\left\{1;-1;5;-5\right\}\)
*Trường hợp 1:
\(\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x-2=5\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=0\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x-2=-1\\y+1=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-6\end{matrix}\right.\)(thỏa mãn)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x-2=-5\\y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)(thỏa mãn)
Vậy: x∈{3;7;1;-3} và y∈{4;0;-6;-2}
b) (3-x)*(2y+5)=4
\(\Leftrightarrow3-x;2y+5\inƯ\left(4\right)\)
\(\Leftrightarrow3-x;2y+5\in\left\{1;-1;2;-2;4;-4\right\}\)
*Trường hợp 1:
\(\left\{{}\begin{matrix}3-x=1\\2y+5=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\frac{1}{2}\end{matrix}\right.\)(loại)
*Trường hợp 2:
\(\left\{{}\begin{matrix}3-x=4\\2y+5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)
*Trường hợp 3:
\(\left\{{}\begin{matrix}3-x=-1\\2y+5=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\2y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\frac{-9}{2}\end{matrix}\right.\)(loại)
*Trường hợp 4:
\(\left\{{}\begin{matrix}3-x=-4\\2y+5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-3\end{matrix}\right.\)
*Trường hợp 5:
\(\left\{{}\begin{matrix}3-x=2\\2y+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{-3}{2}\end{matrix}\right.\)(loại)
*Trường hợp 6:
\(\left\{{}\begin{matrix}3-x=-2\\2y+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\2y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{-7}{2}\end{matrix}\right.\)(loại)
Vậy: x∈{-1;7} và y∈{-2;-3}
Bài 5:
a) Ta có: \(n+5⋮n-1\)
\(\Leftrightarrow5⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(5\right)\)
\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow n\in\left\{2;0;6;-4\right\}\)(thỏa mãn)
Vậy: \(n\in\left\{2;0;6;-4\right\}\)
b) Ta có: \(2n-3⋮n+4\)
\(\Leftrightarrow-3⋮n+4\)
\(\Leftrightarrow n+4\inƯ\left(-3\right)\)
\(\Leftrightarrow n+4\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow n\in\left\{-3;-5;-1;-7\right\}\)(thỏa mãn)
Vậy: \(n\in\left\{-3;-5;-1;-7\right\}\)
c) Ta có: 3n+4⋮2n-1
\(\Leftrightarrow4⋮2n-1\)
\(\Leftrightarrow2n-1\inƯ\left(4\right)\)
\(\Leftrightarrow2n-1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow2n\in\left\{2;0;3;-1;5;-3\right\}\)
\(\Leftrightarrow n\in\left\{1;0;\frac{3}{2};\frac{-1}{2};\frac{5}{2};\frac{-3}{2}\right\}\)
Vì n∈Z
nên n∈{1;0}
Vậy: n∈{1;0}
Giải các phương trình sau:
a) 2 x − 1 3 + 6 3 x − 1 2 = 2 x + 1 3 + 6 x + 2 3 ;
b) x − 2 2 + 3 − 2 x 2 − 4 x − 4 x − 5 = x + 3 2 ;
c) x − 3 + 2 x − 3 − 1 3 = 3 − x 4 ;
d) x + 4 3 − 1 7 = 2 − x 7 + x 3 + x + 1 .