Ta có:

\(100=2^2.5^2\)

\(200=2^3.5^2\)

\(120=2^3.3.5\)

\(BCNN\left(100;200;120\right)=2^3.3.5^2=8.3.25=600\)

Sửa đề:

Ta có:

\(100=2^2.5^2\)

\(200=2^3.5^2\)

\(300=2^2.3.5^2\)

\(BCNN\left(100;200;300\right)=2^3.3.5^2=600\)

\(#NqHahh\)

Ta có:

\(100=2^2.5^2\)

\(200=2^3.5^2\)

\(300=2^2.3.5^2\)

\(BCNN\left(100;200;300\right)=2^2.2^3.3.5^2=4.8.3.25=2400\)

\(#NqHahh\)

\(90+\left\{20-10+\left[12+78+\left(5^2-15\right)\right]\right\}\)

\(=90+\left\{20-10+\left[12+78+\left(25-15\right)\right]\right\}\)

\(=90+\left\{20-10+\left[12+78+10\right]\right\}\)

\(=90+\left\{20-10+\left[90+10\right]\right\}\)

\(=90+\left\{20-10+100\right\}\)

\(=90+\left\{10+100\right\}\)

\(=90+110\)

\(=200\)

\(#NqHahh\)

Sao ngoặc tròn lại bên ngoài ngoặc vuông ạ? Bạn xem lại đề bài.

Ta có:

\(30=2.3.5\)

\(150=2.3.5^2\)

\(BCNN\left(30;150\right)=2.3.5^2=2.3.25=150\)

\(#NqHahh\)

Ta có:

\(24=2^3.3\)

\(40=2^3.5\)

\(168=2^3.3.7\)

\(BCNN\left(24;40;168\right)=2^3.3.5.7=8.3.5.7=840\)

\(x:0,125+x:10+x\times1,9=202,3\)

\(=>x\times8+x\times0,1+x\times1,9=202,3\)

\(=>x\times\left(8+0,1+1,9\right)=202,3\)

\(=>x\times10=202,3\)

\(=>x=202,3:10=20,23\)

Vậy...

\(#NqHahh\)

Số tuổi của con năm nay là:

\(45:3=15\left(tuổi\right)\)

Số tuổi của bố năm con sinh ra là:

\(45-15=30\left(tuổi\right)\)

Đáp số:...

\(#NqHahh\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}\)

\(A=\dfrac{100}{100}-\dfrac{1}{100}\)

\(A=\dfrac{99}{100}\)

__________________

\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(B=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{2}.\left(1-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{2}.\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{2}.\dfrac{98}{99}\)

\(B=\dfrac{49}{99}\)

\(#NqHahh\)