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c: =>x+2>0
hay x>-2
d: =>-4<=x<=3
e: =>\(x\in\varnothing\)
f: \(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -6\end{matrix}\right.\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
bài 2 tìm x
a,106- ( x+ 7) =9
x+7 = 106 - 9
x+7 = 107
x= 107 - 7
x=100
b, 2 x ( x+ 4) + 5 =65
2 x (x+4) = 65 - 5
2 x (x+4) = 60
x+4 = 60:2
x+4= 30
x= 30 - 4
x=26
c, (16x x -32) x 45=0
16 x X - 32 = 0: 45
16 x X - 32 =0
16 x X = 0 + 32
16 x X = 32
X= 32:16
X=2
d, x+4 x x = 100 : 5
X + 4 x X = 20
(1+4) x X = 20
5 x X = 20
X= 20:5
X=4
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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Bài 1:
a, \(x^2\) +2\(x\) = 0
\(x.\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(x\) \(\in\) {-2; 0}
b, (-2.\(x\)).(-4\(x\)) + 28 = 100
8\(x^2\) + 28 = 100
8\(x^2\) = 100 - 28
8\(x^2\) = 72
\(x^2\) = 72 : 8
\(x^2\) = 9
\(x^2\) = 32
|\(x\)| = 3
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(\in\) {-3; 3}
c, 5.\(x\) (-\(x^2\)) + 1 = 6
- 5.\(x^3\) + 1 = 6
5\(x^3\) = 1 - 6
5\(x^3\) = - 5
\(x^3\) = -1
\(x\) = - 1
1) Ta có : a : 3 = 10
a = 10.3
a = 30
2) a) 8x-4x = 1208
=> 4x = 1208
x = 1208 : 4
x = 302
b) 75883 - (31200+x) = 999
31200+x = 75883 - 999
31200+x = 74884
x = 74884 - 31200
x = 43684
3) Ta có : x:5 = 234
x = 234.5
x = 1170
Vậy ta có tập hợp x = {1170}
4) Ta có : Hiệu số tự nhiên lớn nhất là : 321 - 102 = 219
Hiệu số tự nhiên nhỏ nhất là : 103 - 102 = 1
1)a = 3 x 10 = 30
=> a \(\in\)N và a = 30
2) A => x . (8 - 4) = 1208
=> x . 4 = 1208
=> x = 1208 : 4
=> x = 302
1) \(\left(x-4\right)\left(y+1\right)=8\)
Do \(y\)là số tự nhiên nên \(y+1\ge1\)nên
ta có bảng giá trị:
x-4 | 1 | 2 | 4 | 8 |
y+1 | 8 | 4 | 2 | 1 |
x | 5 | 6 | 8 | 12 |
y | 7 | 3 | 1 | 0 |
2) \(\left(2x+3\right)\left(y-2\right)=15\)
Có \(x\)là số tự nhiên nên \(2x+3\ge3\). Ta xét bảng giá trị:
2x+3 | 3 | 5 | 15 |
y-2 | 5 | 3 | 1 |
x | 0 | 1 | 6 |
y | 7 | 9 | 3 |
3) \(xy+2x+y=12\)
\(\Leftrightarrow x\left(y+2\right)+y+2=14\)
\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=14\)
Tiếp tục bạn làm tương tự 1) và 2).
4) \(xy-x-3y=4\)
\(\Leftrightarrow y\left(x-3\right)-x+3=7\)
\(\Leftrightarrow\left(x-3\right)\left(y-1\right)=7\)
Tiếp tục bạn làm tương tự 1) và 2).
a) 2.x+15=27
2.x=27-15
2.x=12
x=12:2
x=6
b)8.x-4.x=1208
x.(8-4)=1208
x=1208:4
x=302
c) (x-5) . 4=0
x-5=0:4
x-5=0
x=0+5
x=5
d)(x-4).(x-3)=0
<=> x-4=0 hoặc x-3=0
x-4=0
x=0+4=4
x-3=0
x=0+3=3
=> (x-4).(x-3)=0
<=> x=4 hoặc x=3
e) x.y-x.2=0
x.(y-2)=0
<=> x=0 hoặc y-2=0
x=0
y-2=0 => x thuộc tập hợp Z
f)75883-(31200+x)=999
31200+x=75883-999
x+31200=74884
x=74884-31200
x=43684
g)(1225+625)-4.x=1000-150=850
4.x=1850-850
4.x=1000
x=1000:4
x=250
chúc bạn học tốt nha
ủng hộ mk với nha
a) 2.X + 15 = 27
2.X = 27 - 15 = 12
X=6
b) 8.X - 4.X = 1208
( 8 - 4 ).X =1208
4.X =1208
X = 302
c) (X - 5 ).4 = 0
X - 5 = 0
X =5
d) ( X - 4 ).( X - 3 ) = 0
Vậy thì X - 4 = 0 hoặc X - 3=0
Vậy X = 4 ; X = 3
e) X.y - 2.X = 0
X.( y - 2 ) = 0
Vậy X=0 hoặc y - 2 = 0
Vậy X=0 và y = 2
f) 75883 - ( 31200 + X ) = 999
31200 + X = 75883 - 999
31200 + X = 74884
X = 43684
g) ( 1225 + 625 ) - 4.X = 1000 - 150
1850 - 4.X = 850
4.X = 1850 - 850
4.X =1000
X = 250