Ngu vcl :))

@Nghiêm đình nam đừng có comment linh tinh ở đây nha ! 

Thick

a: \(A=\dfrac{15}{x+2}+\dfrac{42}{3\left(x+2\right)}=\dfrac{45+42}{3\left(x+2\right)}=\dfrac{29}{x+2}\)

b: Để A là số nguyên thì \(x+2\in\left\{1;-1;29;-29\right\}\)

hay \(x\in\left\{-1;-3;27;-31\right\}\)

\(\text{- ( 2789 _ 435 ) + ( 1789 _ 1435 )}\)

\(=-2789+435+1789-1435\)

\(=\left(-2789+1789\right)+\left(435-1435\right)\)

\(=-1000+-1000\)

\(=-2000\)

\(=-\left(-2010\right)+36.41-36.\left(-59\right)\)

\(=2010+36.\left(41+59\right)\)

\(=2010+36.100\)

\(=2010+3600\)

\(=5610\)

\(-75.\left(18-65\right)-65.\left(75-18\right)\)

\(=-75.18+75.65-65.75+65.18\)

\(=18.\left(-75+65\right)+75.\left(65-65\right)\)

\(=18.\left(-10\right)+75.0\)

\(=-180\)

\(-15:x=3\)

\(x=-15:3\)

\(x=-5\)

\(-3x+8=7\)

\(-3x=-1\)

\(x=\frac{1}{3}\)

\(\left(x-6\right).\left(7-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)

\(\Rightarrow x\in\left\{6;7\right\}\)

\(2.\left(x-3\right)-3.\left(x-5\right)=4.\left(3-x\right)-18\)

\(2x-6-3x+15=12-4x-18\)

\(2x-3x+4x=12-18-15+6\)

\(3x=-15\)

\(\Rightarrow x=-5\)

\(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)

\(-a.c+a.d-d.a+-d.c=-c.\left(a+d\right)\)

\(-c.\left(a+d\right)+a.\left(d-d\right)=-c.\left(a+d\right)\)

\(-c.\left(a+d\right)+a.0=-c.\left(a+d\right)\)

\(\Rightarrow-c.\left(a+d\right)=-c.\left(a+d\right)\)

(3a+2).(2a–1)+(3–a).(6a+2)–17.(a–1)

=6a²−3a+4a−2+18a+6−6a²−2a−17a+17

=(6a²−6a²)+(−3a+4a+18a−2a−17a)+(17−2+6)

=0+0+21

=21

học tốt

a: \(A=\dfrac{1}{15}\cdot\dfrac{225}{x+2}+\dfrac{3}{14}\cdot\dfrac{196}{3x+6}\)

\(=\dfrac{15}{x+2}+\dfrac{14}{x+2}=\dfrac{29}{x+2}\)

b: Để A là số nguyên thì \(x+2\in\left\{1;-1;29;-29\right\}\)

hay \(x\in\left\{-1;-3;27;-31\right\}\)