tim x biet x+10=[x]-10
tim x biet x+10 la boi cua x+1
tim x biet 2x+1 la uoc cua x+82
Bạn Nguyễn Đoan Hạnh cho mình bổ sung nhé
Ư(9)={+-1;+-3;+-9}
Nếu x+1=-1 => x=-2
Nếu x+1=-3 => x = -4
Nếu X+1=-9 => x = -10
x+10 la boi cua x+1
suy ra (x+1)+9 la boi cua x+1
suy ra 9 la boi cua x+1
U(9)={1;3;9}
Neu x+1=1 thi x=0
Neu x+1=3 thi x=2
Neu x+1=9 thi x=8
Vay x thuoc {0;2;8}
Tim x biet:
|x-10|10+|x-11|10=1
cho b E z . Tim so nguyen x, biet b - x = -10 thi :
A. x= -10 + b B. x= -10 - b C. b= -10 -x D. x= b + 10
tim x biet x *5,2- x=4,2 *10
\(x\) \(\times\) 5,2 - \(x\) = 4,2 \(\times\) 10
\(x\) \(\times\) 5,2 - \(x\) \(\times\) 1 = 42
\(x\) \(\times\) ( 5,2 - 1) = 42
\(x\) \(\times\) 4,2 = 42
\(x\) = 42 : 4,2
\(x\) = 10
tim x biet (x+2)/10^10+(x+2)/11^11=(x+2)/12^12+(x+2)/13^13
=>
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=>\(\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}=0\right)\)
vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}=>\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=>x+2=0
=>x=-2
tick nhé
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(=>\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
\(=>\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}\)
Nên \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)khác 0
Suy ra: x+2=0
x =0 - 2
x = -2
TICK NHA
tim x biet :3/[x+2][x+5]+5[x+5][x+10]+7/[x+10][x+15]=x/[x+12][x+17]
tim x biet x +5 =10
tim x biet : 25 - x = 10
bằng 25 - 10 = 15 nha bạn . mình tin chắc bạn sẽ
tim x biet x+ 10 = 12