Câu C nữa em

B. 0 = (x - 2)²(x + 2)

C. (x - 2)²(x + 2) = 2(x + 2)

2(x - 3) - 7 = 5 + (2x - 18)

⇔ 2x - 6 - 7 = 5 + 2x - 18

⇔ 2x - 2x = 5 - 18 + 6 + 7

⇔ 0x = 0 (luôn đúng với mọi x)

Vậy S = R

5,7 × (179 - 130) + (3,4 + 2,3) × (32 + 19)

= 5,7 × 49 + 5,7 × 51

= 5,7 × (49 + 51)

= 5,7 × 100

= 5700

5,7 × (179 - 130) + (3,4 + 2,3) × (32 + 19)

= 5,7 × 49 + 5,7 × 51

= 5,7 × (49 + 51)

= 5,7 × 100

= 5700

\(-11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)-1\dfrac{2}{3}\)

\(=-\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)-\dfrac{5}{3}\)

\(=-\dfrac{47}{4}-\dfrac{41}{6}+\dfrac{9}{2}-\dfrac{5}{3}\)

\(=\left(-\dfrac{47}{4}+\dfrac{9}{2}\right)-\left(\dfrac{41}{6}+\dfrac{5}{3}\right)\)

\(=-\dfrac{29}{4}-\dfrac{17}{2}\)

\(=-\dfrac{63}{4}\)

ĐKXĐ: \(m\ne1\)

Gọi \(\left(d'\right):y+2x-3=0\)

\(\Leftrightarrow\left(d'\right):y=-2x+3\)

Để \(\left(d\right)\perp\left(d'\right)\) thì: \(\left(m-1\right).\left(-2\right)=-1\)

\(\Leftrightarrow-2m+2=-1\)

\(\Leftrightarrow-2m=-3\)

\(\Leftrightarrow m=\dfrac{3}{2}\) (nhận)

\(\Rightarrow\left(d\right):y=\dfrac{1}{2}x+n+2\)

Thay tọa độ điểm A(2; 4) vào (d) ta được:

\(4=\dfrac{1}{2}.2+n+2\)

\(\Leftrightarrow1+n+2=4\)

\(\Leftrightarrow n=4-1-2\)

\(\Leftrightarrow n=1\)

Vậy \(m=\dfrac{3}{2};n=1\)