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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
=27/51/59-7/51/59+1/3
=(27/51/59-7/51/59)+1/3
=20+1/3
=20/1/3
=31/6/13+5/9/41+(-36/6/13)
=(31/6/13+-36/6/13)+5/9/41
=-5+5/9/41
=9/41
duyet nha
a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)
Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)
b: 51/59<1<513/293
c: 31/157<31/155=1/5
17/83>17/85=1/5
Do đó: 31/157<17/83
a)
\( 49.(51 - 4) - 51.(49 + 4)\)
\(=49.51-49.4-51.49-51.4\)
\(=49.(51-4-51)-51.4\)
\(=49.(-4)-51.4\)
\(=4.(-49-51)\)
\(=4.(-100)\)
\(=-400\)
b)
\( 71.64 + 32.(-7) - 13.32\)
\(=71.32.2+32.(-7)-13.32\)
\(=142.32+32.(-7)-13.32\)
\(=32.(142-7-13)\)
\(=32.122\)
\(=3904\)
c)
\( 11 + (-13) + 15 + (-17) + ... + 59 + (-61)\)
\(=(-2)+(-2)+...+(-2)\)
\(=(-2).13\)
\(=-26\)
a. 49.51- 49.4 -51.49 -51.4
= -49.4 - 51.4
=4.(-49-51)
=4.-100
=-400
b.71.64 + 32.(-7) -13.32
= 71.64 + 32.(-7-13)
=71.64 + 32.-20
=71.32.2 +32.-20
=32.(71x2-20)
=32. 122
= 3904
c. (11-13)+(15-17)+..+(59-61)
=-2 x 13
=-26
ngu cái này tao cũng ko bt
ê đừng nói tục nha bẹn