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ĐKXĐ: x ≠ 1;x ≠ 2;x ≠ 3;x ≠ 4
Phương trình



Vậy tập nghiệm của phương trình là S = { 0;5/2 }
a: \(\overline{xx}+\overline{x}+5=125\)
=>10x+x+x=120
=>12x=120
=>x=10
b: \(\overline{xxx}+\overline{xx}+x+x=992\)
=>100x+10x+x+10x+x+2x=992
=>124x=992
=>x=8(nhận)
c: \(4725+\overline{xxx}+\overline{xx}+\overline{x}=54909\)
=>100x+10x+x+10x+x+x=54909-4725
=>123x=50184
=>x=408
d: \(\overline{xxx}-\overline{xx}-\overline{x}-25=4430\)
=>100x+10x+x-10x-x-x=4430+25
=>99x=4425
=>x=4425/99(loại)
e: \(\overline{xxx}+\overline{xx}+\overline{x}+\overline{x}+\overline{x}+1=1001\)
=>111x+11x+3x=1000
=>125x=1000
=>x=8
f: \(35655-\overline{xxx}-\overline{xx}-\overline{x}=5274\)
=>35655-111x-11x-x=5274
=>35655-123x=5274
=>123x=35655-5274=30381
=>x=247
3,455 \(\times\) \(x\) = 345,5
\(x\) = 345,5 : 3,455
\(x\) = 100
ĐKXĐ: x ≠ 3, x ≠ 4, x ≠ 5, x ≠ 6.
Ta có:







Vậy phương trình đã cho có nghiệm là x = 0;x = 9/2.
HD các số hạng x+2 ; x+4 ;........ ; x+1996 lập thành 1
dãy số cách đều với khoảng cách là 2.
Từ x+2 đến x+1996 có: (1996-2) : 2+ 1=998(số hạng)
Tổng các số vế trái :
( x+2) +(x+4) +.....+(x+1996)= ( x+2 +x+1996) x998 : 2=
(2x x +1998) x998 : 2
Vậy ta có ( 2 x x+1998) x998 :2 =1998
tk nha bạn
thank you bạn
(^_^)


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