tìm x
3x-8=17
15-(27-x)=-9
-12+(x-7)=-34
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a) x - (17 - 8) = 5 + (10 - 3x)
x - 17 + 8 = 5 + 10 - 3x
x + 3x = 5 + 10 + 17 - 8
4x = 24
x = 24 : 4
x = 6
b) 25 - (30 + x) = x - (27 - 8)
25 - 30 - x = x - 27 + 8
-x - x = -27 + 8 - 25 + 30
-2x = -14
x = (-14) : (-2)
x = 7
c) (x - 12) - 15 = (20 - 7) - (18 + x)
x - 12 - 15 = 20 - 7 -18 - x
x + x = 20 - 7 - 18 + 12 + 15
2x = 22
x = 22 : 2
x = 11
d) 9 - 25 = (7 - x) - (25 + 7)
9 - 25 = 7 - x - 25 - 7
x = 7 - 25 - 7 - 9 +25
x = -9
a: =>3x=10+14=24
=>x=8
b: =>25-30-x=-19
=>-5-x=-19
=>x+5=19
=>x=14
c: =>15-x=12+14=26
=>x=-11
d: =>4x+11-x-34=67
=>3x-23=67
=>3x=90
=>x=30
e: =>-x-14=289-36-289=-36
=>x+14=36
=>x=22
a) 3x=24
x=8
b) 25-30-x=-19
25-30+19=x
14=x
c) 15-x=26
x=-11
d) 4x+11-x-34=67
3x=90
x=30
e) -14-x= -36
x=-14 -36
x= 22
a/ Ta có :
\(\left|x-2\right|=\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge\left|x+2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge2\)
Dấu "=" xảy ra khi :
\(x\left(x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow2\le x\le2\Leftrightarrow x=2\)
Vậy \(x=2\)
b/ \(\left|2x-1\right|+\left|9-2x\right|\ge\left|2x-1+9-2x\right|\)
\(\Leftrightarrow\left|2x-1\right|+\left|9-2x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(2x-1\right)\left(9-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\9-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\9-2x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le\dfrac{9}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{9}{2}\)
Vậy ....
c/ Ta có : \(\left|3x-20\right|=\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|=\left|3x+7\right|+\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge\left|3x+7+20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge27\)
Dấu "=" xảy ra khi :
\(\left(3x+7\right)\left(20-3x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+7\ge0\\20-3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+7\le0\\20-3x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{7}{3}\\x\le\dfrac{20}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{7}{3}\\x\ge\dfrac{20}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{20}{3}\le x\le-\dfrac{7}{3}\)
Vậy...
d/ \(\left|10-x\right|+\left|x+30\right|\ge\left|10-x+x+30\right|\)
\(\Leftrightarrow\left|10-x\right|+\left|x+30\right|\ge40\)
Dấu "=" xảy ra khi :
\(\left(10-x\right)\left(x+30\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x\ge0\\x+30\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x\le0\\x+30\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10\ge x\\x\ge-30\end{matrix}\right.\\\left\{{}\begin{matrix}10\le x\\x\le-30\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow10\ge x\ge-30\)
Vậy...
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
3x-8=17
x = (17+8):3
x = 25/3
15-(27-x)=-9
x = 27-(15+9)
x = -3
-12+(x-7)= -34
x = (-34+12)+7
x = -15
3x-8 = 17
3x = 17+8
3x = 25
x = 25: 3
x = 25/3
15-( 27-x ) =-9
27-x = 15-(-9)
27-x = 24
x = 27-24
x = 3