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1)
a) 35.23 + 35.41 + 64.65 = 35(23+41) + 64.65 = 35 . 65 + 64.65 =65. 99 = 65(100 - 1 ) = 6500 - 65 = 6435
b) 29.87 - 29.23 + 64.71 = 29(87-23) + 64.71 = 29.64 + 64.71 = 64(71+29) = 64.100= 6400
c) 48.19 + 48.115 + 134.52 = 48(19+115) + 134.52 = 48 . 134+134.52=134(52+48)=134.100=13400
d) 27.121 - 87.27 + 73.34 = 27(121-87) +73 .34 = 27 .34 + 73.34 = 34(73+27)=34.100=3400
e) 125.98-125.46-52.25 = 125(98-46) -52.25 = 125 . 52 - 52.25 =52(125-25) =52.100=5200
f)136.23 + 136.17 - 40.36 = 136(23+17) - 40.36 = 136.40 -40.36 = (136-36)40 = 100.40=4000
g) 17.93 + 116.83 + 17.23 = 17(93+23) + 116.83 = 17.116 + 116.83 =116(83+17) = 116.100=11600
h)19.27 +47.81 + 19.20 = 19(27+20) + 47.81 = 19. 47 + 47.81 = 47(81+19)=47.100 = 4700
i) 87.23 + 13.93 + 70.87 = 87(23+70) + 13.93 = 87 .93 + 13.93 = 93(87+13)=93.100=9300
2)Tìm x
a) [(6x-39):37].4 = 12
(6x - 39) : 37 = 12 : 4 = 3
x =(3.37 + 39): 6 = 25
b) ( x : 3 - 4 ) .5=15
x : 3 -4 = 3
x = (3+4).3 = 21
c) 128 - 3(x+4) =23
3x + 12 = 128 -23 = 105
x = (105-12):3 = 31
d) (3x - 34).73=2.74
3x - 81 = 2.74:73=2.7=14
x =(81 + 14 ) : 3= \(31\frac{2}{3}\)
e) x - [42 + (-28) ] = -8
x - 14 = -8
x- 14 + 14 = -8 + 14
x = 6
1. Tính nhanh:
a, 35.23+35.41+64.65
= 35. (23+41) +64.65
= 35. 64 + 64.65
= 64. ( 35 + 65 )
= 64 . 100
= 6400
b, 29.87 - 29.23 + 64.71
= 29.(87-23) + 64.71
= 29.64 + 64.71
= 64.(29 + 71)
= 64.100
= 6400
c, 48.19 + 48.115 + 134.52
= 48.(19 + 115) + 134.52
= 48.134 + 134.52
= 134.(48 + 52)
= 134.100
= 13400
d, 27.121 - 87.27 + 73.34
= 27.(121 - 87) + 73.34
= 27.34 + 73.34
= 34.(27 + 73)
= 34.100
= 3400
e, 125.98 - 125.46 - 52.25
= 125.(98 - 46) - 52.25
= 125.52 - 52.25
= 52.(125 - 25)
= 52.100
= 5200
f, 136.23 + 136.17 - 40.36
= 136.(23 + 17) - 40.36
= 136.40 - 40.36
= 40.(136 - 36)
= 40.100
= 4000
g, 17.93 + 116.83 + 17.23
= 17.(93+23) + 116.83
= 17.116 + 116.83
= 116.(17 + 83)
= 116.100
= 11600
Lát nữa mình làm tiếp, giờ mình phải đi học rồi
a; |\(x+2\)| = 0
\(x+2=0\)
\(x\) = - 2
Vậy \(x\) = - 2
b; |\(x-5\)| = |-7|
| \(x-5\) | = 7
\(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5+7\\x=-7+5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
Vậy \(x=12\)
\(x=-2\)
a: \(\Leftrightarrow2x-2+4x+8=-12\)
=>6x+6=-12
=>6x=-18
hay x=-3
b: \(\Leftrightarrow-10x-15-12+9x=13\)
=>-x-27=13
=>-x=40
hay x=-40
c: \(\Leftrightarrow-10x+70+20-5x=-15\)
\(\Leftrightarrow-15x=-105\)
hay x=7
d: \(\Leftrightarrow8x-12-7x+14=10\)
=>x+2=10
hay x=8
e: \(\Leftrightarrow-12x-18+14x+2=2\)
=>2x-16=2
hay x=9
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
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!!
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
d, \(\left(3x-2^4\right).7^3=2.7^4\)
\(\Rightarrow3x-2^4=2.7^4:7^3\)
\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)
Vậy.....
e, \(x-\left[42+\left(-28\right)\right]=-8\)
\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)
Vậy.....
g, \(x-7=-5\)
\(\Rightarrow x=-5+7\Rightarrow x=2\)
Vậy.....
h, \(15-5\left(x+4\right)=-12-3\)
\(\Rightarrow15-5x-20=-15\)
\(\Rightarrow-5x=-15-15+20\)
\(\Rightarrow-5x=-10\Rightarrow x=2\)
Vậy.....
Chúc bạn học tốt!!!
d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)
\(\Rightarrow3x=14+16=30\)
\(\Rightarrow x=\dfrac{30}{3}=10\)
e/ Đễ ==> tự lm thì tốt hơn nhé
g/ Đễ ==> tự lm thì tốt hơn nhé
h/ \(15-5\left(x+4\right)=-12-3\)
\(\Rightarrow15-5x-20=-15\)
\(\Rightarrow-5x=-15+20-15=-10\)
\(\Rightarrow x=\dfrac{-10}{-5}=2\)
i/ \(\left(7-x\right)-\left(25+7\right)=-25\)
\(\Rightarrow7-x-25-7=-25\)
\(\Rightarrow-x=-25-7+7+25\)
\(\Rightarrow-x=0\Rightarrow x=0\)
k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)
l/ \(\left|x-3\right|=7-\left(-2\right)\)
\(\Rightarrow\left|x-3\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)