95-105:X=60
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a) Ta có \(\dfrac{72}{95}\times\dfrac{105}{105}=\dfrac{72}{95}\times1=\dfrac{72}{95}=\dfrac{72}{95}\)
Vậy \(\dfrac{72}{95}=\dfrac{72}{95}\times\dfrac{105}{105}\)
b) Ta có \(\dfrac{72}{95}\times\dfrac{99}{98}=\dfrac{72}{95}\times\left(1+\dfrac{1}{98}\right)=\dfrac{72}{95}\times1+\dfrac{72}{95}\times\dfrac{1}{98}=\dfrac{72}{95}+\dfrac{72}{95}\times\dfrac{1}{98}>\dfrac{72}{95}\)Vậy \(\dfrac{72}{95}< \dfrac{72}{95}\times\dfrac{99}{98}\)
a) 72/95 ... 72/95 x 105/105
72/95 ... 72/95 x 1 ( 105/105 = 1 )
72/95 = 72/95 x 105/105
b) 72/95 ... 72/95 x 99/98
72/95 x 1 ... 72/95 x 99/98 ( 99/98 > 1 )
72/95 < 72/95 x 99/98
1) 9 . (x + 7) - 12 = 24
9 . (x + 7) = 24 + 12
9 . (x + 7) = 36
x + 7 = 36 : 9
x + 7 = 4
x = 4 - 7
x = -3
2) 12 - 3x = -30
3x = 12 - (-30)
3x = 12 + 30
3x = 42
x = 42 : 3
x = 14
3) 95 - 105 : x = 60
105 : x = 95 - 60
105 : x = 35
x = 105 : 35
x = 3
4) x + 35 = 12
x = 12 - 35
x = -23
5) (-24) - (10 - x) = 43
-24 - 10 + x = 43
-34 + x = 43
x = 43 - (-34)
x = 43 + 34
x = 77
6) 6 - (17 + x) = -16
6 - 17 - x = -16
-11 - x = -16
x = -11 - (-16)
x = -11 + 16
x = 5
7) (x - 18) - (-3) = 0
x - 18 + 3 = 0
x - 18 = 0 - 3
x - 18 = -3
x = -3 + 18
x = 15
8) 25 - (x - 6) = -1
25 - x + 6 = -1
25 - x = -1 - 6
25 - x = -7
x = 25 - (-7)
x = 25 + 7
x = 32
1)9.(x+7)-12=24
9.(x+7)=24+12
9.(x+7)=36
x+7=36:9
x+7=4
x=4-7
x=-3
\(a.83-\left(-756-17\right)+\left(50-756\right)=83+756+17+50-756=\left(83+17\right)+50+\left(756-756\right)=100+50=150\)
\(b.-105+\left(-34-95\right)-166=-105+\left(-34\right)-95-166=\left(-105-95\right)-\left(34+166\right)=\left(-200\right)-200=-400\)
\(c,-\left(39+228-407\right)+\left(118-161\right)=-39-228+407+118-161=\left(-39-161\right)-\left(228-118\right)+407=\left(-200\right)-110=-310\)
\(d,5^{10}:5^8+60:12+\left(-10\right)=5^2+5+\left(-10\right)=5\left(5+1-2\right)=5.4=20\)
\(e,-342-\left(161-342\right)-39=-342-161+342-39=\left(-342+342\right)-\left(161+39\right)=-200\)\(g,7^5:7^3+\left(-187-149\right)-213=7^2+\left(-187\right)-149-213=\left(49-149\right)-\left(187+213\right)=\left(-100\right)-400=-500\)
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
Sửa đề: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
Ta có: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x-105=0\)
\(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1+\frac{x-7}{93}-1+\frac{x-95}{5}-1+x-100=0\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}+\frac{x-100}{93}+\frac{x-100}{5}+\frac{x-100}{1}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\right)=0\)
mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\ne0\)
nên x-100=0
hay x=100
Vậy: x=100
\(95-105:x=60\)
\(105:x=95-60\)
\(105:x=35\)
\(x=105:35\)
\(x=3\)
Ko ghi lại đề!!
=> 105 : X = 35
=> X = 105/35 =3
Vậy...