Tính giúp mình nha.
\(H=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03\)
\(I=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
Ai làm đúng thì mình tick cho.
Tính
I=10101.(\(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\))
H=(9\(\frac{30303}{80808}+7\frac{303030}{484848}\))+4,03
I=10101(5/111111+5/222222+4/3.7.11.13.37)
I=10101(5/111111+5/222222+4/111111)
I=10101(10/222222+5/222222+8/222222)
I=10101.23/222222
I=232323/222222
\(I=\frac{7}{22}\)
\(H=\frac{2103}{100}=21,03\)
H=(9 30303 phần 80808+7 303030 phần 484848)+4,03
I=10101.(5 phần 111111+5 phần 222222-4 phần 3.7.11.13.37)
B=71\3\45 - (43\8\45 - 1\17\57)
H=( 9\30303\80808 + 7\303030\484848 )+4,03
I=10101.(5\111111 + 5\222222 - 4\3.7.11.13.37)
H=(9 30303/80808+7 303030/484848)+4.03
I=10101.(5/111111+5/222222-4/3.7.11.13.37)
giải thích đầy đủ ạ
H=\(\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03\)
\(H=\left(9+\dfrac{3}{8}+7+\dfrac{5}{8}\right)+4.03=17+4.03=21.03\)
Tính hợp lý giá trị biểu thức sau:
H={\(9\frac{30303}{80808}\)+7\(\frac{303030}{484848}\)}+4,03 I =10101.{\(\frac{5}{111111}\)+\(\frac{5}{222222}\)-\(\frac{4}{3.7.11.13.37}\)}
giúp mình nha các bạn! siêu gấp lắm@@@
Mình like cho ...
\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
tính hợp lí
a)\(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}\)
b)\(H=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03\)
\(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}=\left(\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\right).\frac{15}{78}=\left(\frac{39}{4}.40\right).\frac{15}{78}=390.\frac{15}{78}=75\)\(H=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03=\left(9\frac{3}{8}+7\frac{30}{48}\right)+4,03=17+4,03=21,03\)
\(I=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
B= 10101.\(\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(B=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\left(\dfrac{2}{222222}+\dfrac{5}{222222}\right)\)
\(=10101\cdot\dfrac{1}{31746}=\dfrac{7}{22}\)