Ta có: \(A=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{352}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{352-2852-5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{-835}{4641}+\dfrac{8}{39}\)

\(=\dfrac{3}{119}\)

Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\)

Ta có:\(3\dfrac{1}{117}.\dfrac{1}{119}-\dfrac{4}{117}.5\dfrac{118}{119}-\dfrac{5}{117.119}+\dfrac{8}{39}\)

=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+24a\)

\(\text{ }\)=3b+ab-24a+4ab-5ab+24a

\(=3b=3.\dfrac{1}{119}=\dfrac{3}{119}\)

\(N=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{1}{39}+\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{\dfrac{117119}{1000}}+\dfrac{8}{39}\)

\(=\dfrac{1}{4641}-\dfrac{2852}{13923}-\dfrac{5000}{117119}+\dfrac{8}{39}\)

\(=-\dfrac{9827881}{232949691}\)

Đặt \(\dfrac{1}{117}=x;\dfrac{1}{119}=y\)

\(\Rightarrow\dfrac{1}{39}=3x\)

Ta có: \(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+8.3x\)

\(=3y+xy-20x-4x+4xy-5xy+24x\)

\(=3y\)

Thay \(y=\dfrac{1}{119}\rightarrow A:\)

\(A=3.\dfrac{1}{119}=\dfrac{3}{119}\)

Vậy \(A=\dfrac{3}{119}.\)

Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) thay vào A được:

A=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+\dfrac{8}{39}\)

=\(3b+ab-24a+4ab-5ab+\dfrac{8}{39}\)

=\(3b-24a+\dfrac{8}{39}\) (1)

Thay \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) vào (1) ta đuợc:

A=\(\dfrac{3}{119}-\dfrac{24}{117}+\dfrac{8}{39}=\dfrac{3}{119}-0=\dfrac{3}{119}\)

Chúc các bn học tốt

\(\dfrac{3}{119}\)

Đặt : \(\dfrac{1}{117}\) = x ; \(\dfrac{1}{119}\) = y .

A = ( 3 + x)( 4 + y) - (1 + 1 - x)(5 + 1 - y) - 5y

<=> A = 12 + 3y + 4x + xy - ( 2 - x)( 6 - y) - 5y

<=> A = 12 + 3y + 4x + xy - 12 + 2y + 6x - xy - 5y

<=> A = 10x

<=> A = \(\dfrac{10}{117}\).

Vậy A = \(\dfrac{10}{117}\)

Sửa đề: \(C=3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}+\dfrac{5}{119}-\dfrac{10}{117}\)

\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(1+1-\dfrac{1}{117}\right)\left(5+1-\dfrac{1}{110}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)

\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(2-\dfrac{1}{117}\right)\left(6-\dfrac{1}{119}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)

Đặt \(a=\dfrac{1}{117}\) và \(b=\dfrac{1}{119}\) ta có:

\(C=\left(3+a\right).\left(4+b\right)-\left(2-a\right)\left(6-b\right)+5b-10a\)

\(=12+3b+4a+ab-12+2b+6a-ab+5b-10a\)

\(=10b=10.\dfrac{1}{119}=\dfrac{10}{119}\)

10\119

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)