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a) 105*x-20-18-16-...-2=19+17+15+...+3+1
105*x=1+2+3+4+...+20
105*x=\(\frac{\left(20+1\right)x20}{2}\)
105*x=210
x=2
Vậy x = 2
b) 1-6+11-16+21-26+...+91-96+101
=101-96+91-86+...+21-16+11-6+1
=(101-96)+(91-86)+...+(21-16)+(11-6)+1
=5+5+...+5+5+1 ( có (101-6):5+1);2=10 số 5)
=5x10+1=51
a)105xX-20-18-16-...-2=19+17+15+...+3+1
105xX-(20+18+16+...+2)=19+17+15+...+3+1
105xX-110=100
105xX=100+110
105xX=210
X=210:105
X=2
Tích trên có tận cùng là 7 c/s 0. Ai thấy mình đúng thì chọn nhé!
Bài 1
a; \(\dfrac{7}{19}\) x \(\dfrac{1}{3}\) + \(\dfrac{7}{19}\) x \(\dfrac{2}{3}\)
= \(\dfrac{7}{19}\) x (\(\dfrac{1}{3}+\dfrac{2}{3}\))
= \(\dfrac{7}{19}\) x 1
= \(\dfrac{7}{19}\)
b; 15 x \(\dfrac{2121}{4343}\) + 15 x \(\dfrac{212121}{434343}\)
= 15 x \(\dfrac{21}{43}\) + 15 x \(\dfrac{21}{43}\)
= 15 x \(\dfrac{21}{43}\) x (1 + 1)
= 15 x \(\dfrac{21}{43}\) x 2
= (15 x 2) x \(\dfrac{21}{43}\)
= 30 x \(\dfrac{21}{43}\)
= \(\dfrac{630}{43}\)
26,42 : x - 19,04 : x = 3,6
( 26,42 - 19,04 ) : x = 3,6
7,38 : x = 3,6
x = 7,38 : 3,6
x = 2,05
~ Hok tốt ~
26,42 : x - 19,04 : =3,6
x nhân 3,6=26,42 - 19,04
x nhân 3,6 = 7,38
x= 7,38 : 3,6
x= 2,05
Bạn ơi dấu . là dấu nhân nhé!!!
a. 15 -(x - 8) = 2
(x - 8)=15-2
(x - 8)=13
x =13+8
x=21
b. 2.x-19=31
2x =31+19
2x =50
x =50:2
x =25
c. 30-3.(x+1)=15
3.(x+1)=30-15
3.(x+1)=15
(x+1)=15:3
(x+1)=5
x =5-1
x=4
T mk nhé bạn ^...^ ^_^
1) 31-3.(x+2) = 7
3.(x+2) = 31- 7
3.(x+2) = 24
x+2 = 24:3
x + 2 = 8
x = 8-2
x = 6
2) 175-25.(2x-1) = 175
25.(2x-1) = 175- 175 = 0
2x-1 = 0 : 25 = 0
2x = 0 + 1 = 1
x = 1 : 2 = 0,5
3) 25 - x/2 = 19
x/2 = 25-19
x/2 = 6
x : 2 = 6
x = 6 x 2 = 12
4) 2.(x-1)+3 = 15
2.(x-1) = 15 - 3 = 12
x - 1 = 12 : 2 = 6
x = 6+1 = 7
A) x/13+15/26=46/52
x/13+15/26=23/26
x/13=23/26-15/26
x/13=4/13
x=4
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
`@` `\text {Ans}`
`\downarrow`
Lớp 6 - 8 phải k ạ? Lớp 5 chưa học dạng ntnay.
\(15 . ( x- 3 ) + 3 (x+ 26 ) = 19 . ( x + 1)\)
`=>`\(15x-45+3x+78=19x+19\)
`=> 15x - 45 + 3x + 78 - 19x - 19 = 0`
`=> (15x + 3x - 19x) + (-45 + 78 - 19) = 0`
`=> -x + 14 = 0`
`=> -x = -14`
`=> x = 14`
Vậy, `x = 14.`