a) \(A=\left(\dfrac{1-x^3}{1-x}-x\right):\dfrac{1-x^2}{1-x-x^2+x^3}\)

\(A=\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\dfrac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)

\(A=\left(x^2+1\right)\cdot\left(1-x\right)\)

b)

Thay x = \(-1\dfrac{2}{3}=-\dfrac{5}{3}\) vào A có:

\(A=\left(\left(-\dfrac{5}{3}\right)^2+1\right)\cdot\left(1-\left(-\dfrac{5}{3}\right)\right)=\dfrac{272}{27}\)

c)

\(A< 0\Leftrightarrow\left(x^2+1\right)\left(1-x\right)< 0\)

mà \(x^2+1>0\forall x\in R\) \(\Rightarrow1-x< 0\Leftrightarrow x>1\)

\(\dfrac{x^2+1}{x+1}+\dfrac{x^2+2}{x-2}=-2\) (ĐK: \(x\ne-1;x\ne2\))

\(\Leftrightarrow\dfrac{\left(x^2+1\right)\left(x-2\right)+\left(x^2+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=-2\)

\(\Leftrightarrow\dfrac{x^3-2x^2+x-2+x^3+x^2+2x+2}{x^2-x-2}=-2\)

\(\Leftrightarrow2x^3-x^2+3x=-2x^2+2x+4\)

\(\Leftrightarrow2x^3+x^2+x-4=0\)

Bấm máy tính, ta có: x = 1 TMĐK

Vậy \(S=\left\{1\right\}\)

a)

Áp dụng ĐL cosin cho \(\Delta ABC\) có:

\(BC^2=AB^2+AC^2-2AB\cdot AC\cdot cos120^o\)

\(BC=\sqrt{3^2+4^2-2\cdot3\cdot4\cdot\left(-\dfrac{1}{2}\right)}=\sqrt{37}\)

b)

Áp dụng hệ quả ĐL cosin cho \(\Delta ABC\) có:

\(cosB=\dfrac{AB^2+BC^2-AC^2}{2\cdot AB\cdot AC}\)

\(cosB=\dfrac{3^2+\sqrt{37}^2-4^2}{2\cdot3\cdot\sqrt{37}}=\dfrac{5\sqrt{37}}{37}\)

\(\Rightarrow\widehat{B}\approx35^o\)

\(\Rightarrow\widehat{C}=180^o-120^o-35^o=25^o\)

c)

\(S_{ABC}=\dfrac{1}{2}AB\cdot AC\cdot sin120^o=\dfrac{1}{2}\cdot3\cdot4\cdot\dfrac{\sqrt{3}}{2}=3\sqrt{3}\) 

1.B  2.D  3.B  4.C  5.D

1.B  2.B  3.B  4.A  5.B

1.B  2.A  3.C  4.B

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1.C  2.D  3.B  4.B  5.D  6.B  7.C