x-4*25%+10=200
giup mik
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25 – 2x = 16 – 3x
-2x + 3x = 16 - 25
x = -9
Vậy x = -9
x - (47 - 22 ) = 5 + (10-4x)
x - 47 + 22 = 5 + 10 - 4x
x + 4x = 5 + 10 - 22 + 47
5x = 40
x = 40 : 5
x = 8
Vậy x = 8.
~ HOK TỐT ~
\(\text{25 – 2x = 16 – 3x}\)
\(2x+3x=16-25\)
\(5x=-9\)
\(\Rightarrow x=\frac{-9}{5}\)
\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)
\(x-47+22=5+10-4x\)
\(x+4x=5+10-22+47\)
\(5x=40\)
\(\Rightarrow x=8\)
học tốt
a, 2\(x\) + 4 - 5\(x\) = -11
-(5\(x\) - 2\(x\)) + 4 = -11
-3\(x\) + 4 = -11
3\(x\) = 11 + 4
3\(x\) = 15
\(x\) = 15 : 3
\(x\) = 5
b, \(x\) - (-5) = 8
\(x\) + 5 = 8
\(x\) = 8 - 5
\(x\) = 3
1/(-37) +14+26+37
=[(-37)+37] +(14+26)
= 0 + 40
= 40
2/(-24) +6 +10+24
=[(-24)+24] +(10+6)
= 0 +16
=16
3/15+23+9+(-25) +9+(-23)
=[(-23)+23] +[(-25)+15] +(9+9)
= 0 + (-10) +18
= 0 + 8
= 8
4/60 +33 +(-50) +(-33)
=[(-33)+33]+[60 +(-50)]
= 0 + 10
= 10
5/(-16)+ (-209) + (-14) +209
=[(-209)+209] +[(-16)+(-14)]
= 0 +(-30)
= -30
6/(-12) +(-13)+36 +(-11)
=[(-12) +(-11)] +[ 36+(-13)]
= (-23) + 23
= 0
Chúc bn hok tốt !!!
1, (-37)+14+26+37=[(-37)+37]+14+26=0+40=40
2, (-24)+6+10+24=[(-24)+24]+6+10=0+16=16
3 , 15+23+(-25)+(-23)=[15+(-25)]+[23+(-23)]=-10
4, 60+33+(-50)+(-33)= [60+(-50)]+[33+(-33)]=10
5, (-16)+(-209)+(-14)+209=[(-16)+(-14)]+[(-209)+209]=-30
6, (-12)+(-13)+36+(-11)=[(-12)+(-11)]+[(-13)+36]=(-23)+23=0
\(155-\frac{25}{2}\times4\times\frac{1}{10}+x\div\frac{3}{4}=200\)
\(\rightarrow155-5+x\div\frac{3}{4}=200\)
\(\rightarrow150+x\div\frac{3}{4}=200\)
\(\rightarrow x\div\frac{3}{4}=200-150\)
\(\rightarrow x\div\frac{3}{4}=50\)
\(\rightarrow x=50\times\frac{3}{4}\)
\(\rightarrow x=\frac{75}{2}\)
Vậy \(x=\frac{75}{2}\)
a, 17-(2+x)=3
=> x+2= 14
=> x=12
b, (6+x)+(17-21)= -25
6+x- 4= -25
=> x+2=-25
=> x= -27
c, nhận xét / x-1/>=0
=> x-1= 3
=> x=4
Đề bài : \(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)< x< \dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)\)
\(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)=\dfrac{97}{20}\times\dfrac{101}{50}=\dfrac{9797}{1000}\)
\(\dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)=\dfrac{91}{10}\times\dfrac{48}{5}=\dfrac{2184}{25}=\dfrac{2184\times40}{25\times40}=\dfrac{87360}{1000}\)
\(\Rightarrow\dfrac{9797}{1000}< x< \dfrac{87360}{1000}\)
Vậy \(x\in\left(\dfrac{9797}{1000};\dfrac{87360}{1000}\right)\)
\(x-4\times25\%+10=200\)
\(=x-1+10=200\)
\(x-1=200-10\)
\(x-1=190\)
\(x=190+1\)
\(x=191\)