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128 - 3 .(x+4) =23
3.(x+4)=128-23
3.(x+4)=105
x+4=105:3
x+4=35
x=35-4
x=31
(7 - x) -( 25 +7)=-25
(7 -x)-32=-25
7-x=-25+32
7-x=7
x=7-7
x=0
1. 11(x-9)=77
x-9=7
x=16
2. 4(x-3)=48
x-3=12
x=15
3.3(x-8)=81
(x-8)=27
x=35
120 + 25 x 3 = ( x + 23 ) : 4
120 + 75 = ( x + 23 ) : 4
195 x 4 = x + 23
x + 23 = 780
x = 780 - 23 = 757
\(120+25.3=\left(x+23\right):4\)
\(120+75=\left(x+23\right):4\)
\(195.4=x+23\)
\(\Rightarrow780=x+23\)
\(\Rightarrow x=757\)
\(a,\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=\left(-1\right)+1+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}=\dfrac{5}{7}\)
b, \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}.1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}=1\)
a. \(\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=-1+1+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\)
b. \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\cdot1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}\)
\(=1\)
Câu 1 :
a ) (-11)*3 + (-5)-20117-0 -(-10)
= -33+ (-5) -0 - (-10)
= -38-0-(-10)
=-38-(-10)
=-28
b) 25 *(23-17)-25*(23-17)
= 25*6-25*6
=150-150
=0
c) 100[5*(9-11)^2+(-4)^2
=100*[5*(-2)^2 +16
=100*(20+16)
=100*36
=3600
Câu 2:
2*|x-2|=40
|x-2| = 40/2
|x- 2 | = 20
TH1=> x-2 =20=> x= 22
TH2 => x-2 = -20=> x = -20
kb vs mk nha
2)2|x-2|=40=>|x-2|=20=>x-2=20 hoặc x-2=-20=>x=22 hoặc x=-18
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
\(\Rightarrow x=2\)