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a, => 3-17+x=-12
=> x-14=-12
=> x=-12+14 = 2
b, => 26-x+7=0
=> 33-x=0
=> x=33-0=33
c, => 250-2+x=5
=> x+248=5
=> x=5-248 = -243
d, => 30+32-x=10
=> 62-x=10
=> x=62-10=52
Tk mk nha
Bài 1:
a) \(24 - (-15) - 2\)
\(=39-2\)
\(=37\)
b) \((-85) + 10 - (-85) - 50\)
\(=[(-85)-(-85)]+10-50\)
\(=0+10-50\)
\(=10-50\)
\(=-40\)
c) \(71 - (-30) - (+18) + (-30)\)
\(=[(-30)-(-30)]+71-(+18)\)
\(=0+71-18\)
\(=71-18\)
\(=53\)
d) \(-(30) - (+37) + (+37) + (-85)\)
\(=[-(+37)+(+37)]-(30)+(-85)\)
\(=0-(30)+(-85)\)
\(=(-30)+(-85)\)
\(=-115\)
e) \((35-815) - (795-65)\)
\(=(-780)-730\)
\(=-1510\)
g) \((2002-79+15) + (-79+15)\)
\(=1938+(-64)\)
\(=1874\)
Bài 2:
a) \(25 - (30+x) = x - (27-8)\)
\(25-30-x=x-27+8\)
\(x+x=25-30+27-8\)
\(2x=14\)
\(x=14\div2\)
\(x=7\)
b) \((x-12) - 15 = (20-17) - (18+x) \)
\(x-12-15=13-18-x\)
\(x-27=-5-x\)
\(x+x=-5+27\)
\(2x=22\)
\(x=22\div2\)
\(x=11\)
c) \(15 - x = 7 - (-2)\)
\(15-x=9\)
\(x=15-9\)
\(x=6\)
d) \(x - 35 = (-12) - 3\)
\(x-35=-15\)
\(x=-15+35\)
\(x=20\)
e) \(\left|5-x\right|-26=-15\)
\(\left|5-x\right|=-15+26\)
\(\left|5-x\right|=11\)
Từ đây ta có:
*Nếu \(5-x=11\)
\(x=5-11\)
\(x=-6\)
*Nếu \(5-x=-11\)
\(x=5-(-11)\)
\(x=16\)
Vậy \(x=-6;x=16\)
b) -26 - ( x - 7 ) = 0
x - 7 = -26 - 0
x - 7 = -26
x = -26 + 7
x = -19
Bài 2 :
a) 3 - ( 17 - x ) = -12
( 17 - x ) = 3 - ( -12 )
( 17 - x ) = 15
x = 17 - 15
x = 2
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
a ) 2 - x = 17 - ( - 5 )
2 - x = 17 + 5
2 - x = 22
x = 2 - 22
x = - 20
Vậy x = - 20
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
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) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
\(a,3-\left(17-x\right)=-12\\ \Rightarrow17-x=15\\ \Rightarrow x=2\\ b,-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\\ c,25+\left(-2+x\right)=5\\ \Rightarrow-2+x=20\\ \Rightarrow x=18\\ d,30+\left(32-x\right)=10\\ \Rightarrow32-x=-20\\ \Rightarrow x=52\)
a) `3-(17-x)=-12`
`3-17+x=-12`
`x=-12-3+17`
`x=2`
b) `-26-(x-7)=0`
`-26-x+7=0`
`-19-x=0`
`x=-19`
c) `25+(-2+x)=5`
`25-2+x=5`
`x=5-25+2`
`x=-18`
d) `30+(32-x)=10`
`30+32-x=10`
`62-x=10`
`x=52`
a: -26-(x-7)=0
=>x-7=-26
=>x=-26+7=-19
b: \(24-\left(30+x\right)=-7\)
=>\(30+x=24-\left(-7\right)\)
=>\(x+30=24+7=31\)
=>x=31-30=1
c: \(3-\left(17-x\right)=-12\)
=>\(17-x=3-\left(-12\right)\)
=>\(17-x=3+12=15\)
=>x=17-15=2
d: \(\left(5-x\right)-7=-15\)
=>\(5-x=-15+7=-8\)
=>\(x=5-\left(-8\right)=5+8=13\)