\(4+\frac{11}{8}-\frac{5}{6}=\frac{96+33-20}{24}=\frac{109}{24}\)

\(\left(\frac{1}{2}+\frac{4}{7}\right)-\left(\frac{3}{7}-\frac{3}{10}\right)=\frac{1}{2}+\frac{4}{7}-\frac{3}{7}+\frac{3}{10}\)

\(\left(\frac{1}{2}+\frac{3}{10}\right)+\left(\frac{4}{7}-\frac{3}{7}\right)=\frac{4}{5}+\frac{1}{7}=\frac{28+5}{35}=\frac{33}{35}\)

ban Chitanda Eru oi to bao nhung ma nhan voi bao nhieu ma ra duoc 96

b = ( 1+2-3-4) + ( 5+6-7-8) + .............. + (301+302-303-304) + (305 +305)

b = -4 + (-4) + ..................+ (-4) + 610

b= -4 x  76 + 610

b= -304 + 610

b= 306

\(\frac{7}{11}:\frac{3}{5}+\frac{4}{11}x\frac{5}{6}\)

=\(\frac{7}{11}x\frac{5}{3}+\frac{20}{66}\)

=\(\frac{35}{33}+\frac{10}{33}\)

=\(\frac{45}{33}\)

=\(\frac{15}{11}\)

= 7/11 x 5/3 + 4/11 x 5/6

= 35/33 + 10/33

=15/11

CHÚC BẠN HỌC TỐT

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)