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\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
Câu 1:
a; x/7 = - 9/14
x = - 9/14 x 7
x = - 9/2
Vậy x = -9/2
b; - 16/36 = - y/4
-4/9 = - y/4
y = -4/9 x (-4)
y = 16/9
Vậy y = 16/9
Câu 1c:
-5/12 = x/60
x = -5/12 x 60
x = - 25
Vậy x = - 25
d: -4/8 = x/-10 = -7/y
-4/8 = x/-10
x = -4/8 x (-10)
x = 5
-4/8 = -7/y
y = - 7 : (-4/8)
y = 7 x 8/4
y = 14
Vậy (x; y) = (5; 14)
1.Tìm x,y thuộc Z biết
1,x+(-45)=(-62)+17
2,x+29=|-43|+(-43)
3,43+(9-21)=317-(x+317)
4,|x|+|-4|=7
5,|x|+|y|=0
6,(15-x)+(x-12)=7-(-5+x)
7,(2x-5)^2=9
8,(2x+6).(x-9)=0
9,(1-3x)^3=-8
10,3x+4y-xy=15
3.Tìm x+y biết
|x|=5
|x|=7
4.Tìm giá trị lớn nhất hoặc nhỏ nhất của các biểu thức sau (x,y thuộc Z)
A=|x-3|+1
B=3-|x+1|
C=|x-5|+|y+3|+7
\(\text{ x . (x - 3) = 0}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
vậy_______
1 ) 5 - ( 10 - x ) = 7
10 - x = 5 - 7
10 - x = - 2
x = 10 - ( - 2 )
x = 12
Vậy x = 12
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~
Câu 1:
16 - 18|x| = - 8 - 14|x|
18|x| - 14|x| = 16 + 8
4|x| = 24
|x| = 24 : 4
|x| = 6
x = - 6 hoăc x = 6
Vậy x ∈ {-6; 6}
e) 15-3|4x +2| = 3
3|4x + 2| = 15 - 3
3|4x + 2| = 12
|4x + 2| = 12 : 3
|4x + 2| = 4
4x + 2 = -4 hoặc 4x + 2 = 4
Th1: 4x + 2 = - 4
4x = - 4 - 2
4x = - 6
x = - 6/4
x = - 3/2
Th2: 4x + 2 = 4
4x = 4 - 2
4x = 2
x = 2/4
x = 1/2
Vậy x ∈ {- 3/2; 1/2}
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
( 2 x y + 2/15 ) x 3 = 4/5
( 2 x y + 2/15 ) = 4/5 : 3
( 2 x y + 2/15 ) = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 :2
y = 1/15
(2 x y + 2/15) x 3 = 4/5
2 x y + 2/15) = 4/5 : 3
2 x y + 2/15 = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 : 2
y = 1/15
7/9 x (2 - 1/3 x y) = 14/15
(2 - 1/3 x y) = 14/15 : 7/9
(2 - 1/3 x y) = 6/5
2 - y = 6/5 x 1/3
2 - y = 2/5
y = 2/5 + 2
y = 12/5
4/21 + 5 x y - 8/7 = 1/3
4/21 + 5 x y = 1/3 + 8/7
4/21 + 5 x y = 31/21
5 x y = 31/21 - 4/21
5 x y = 9/7
y = 9/7 : 5
y = 9/35
7/12 x y - 3/12 x y = 5
y x (7/12 - 3/12) = 5
y x 1/3 = 5
y = 5 : 1/3
y = 15