\(\) a: ĐKXĐ: \(x{\ne\pm1;x\ne\frac{1}{2}\left.\right.}\)

\(A = \left(\right. \frac{1}{1 - x} + \frac{2}{x + 1} - \frac{5 - x}{1 - x^{2}} \left.\right) : \frac{1 - 2 x}{x^{2} - 1}\)

\(= \left(\right. \frac{- 1}{x - 1} + \frac{2}{x + 1} - \frac{x - 5}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} \left.\right) \cdot \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{- 2 x + 1}\)

\(= \frac{- \left(\right. x + 1 \left.\right) + 2 \left(\right. x - 1 \left.\right) - x + 5}{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} \cdot \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{- 2 x + 1}\)

\(= \frac{- x - 1 + 2 x - 2 - x + 5}{- 2 x + 1} = \frac{2}{- 2 x + 1}\)

\(\) b: Để A>0 thì \(\frac{2}{- 2 x + 1} > 0\)

\(2x-1<0\) vì \(-2<0\)

\(x<\frac12\) (nhận)

vậy \(x<\frac12\) và \(x\ne-1\) thì \(A>0\)

a: \(x^{2} - 3 x + 1 > 2 \left(\right. x - 1 \left.\right) - x \left(\right. 3 - x \left.\right)\)

=>\(x^{2} - 3 x + 1 > 2 x - 2 - 3 x + x^{2}\)

=>-3x+1>-x-2

=>-2x>-3

=>\(x < \frac{3}{2}\)

b: \(\left(\left(\right. x - 1 \left.\right)\right)^{2} + x^{2} < = \left(\left(\right. x + 1 \left.\right)\right)^{2} + \left(\left(\right. x + 2 \left.\right)\right)^{2}\)

=>\(x^{2} - 2 x + 1 + x^{2} < = x^{2} + 2 x + 1 + x^{2} + 4 x + 4\)

=>-2x+1<=6x+5

=>-7x<=4

=>\(x > = - \frac{4}{7}\)

c: 

\(\left(\right. x^{2} + 1 \left.\right) \left(\right. x - 6 \left.\right) < = \left(\left(\right. x - 2 \left.\right)\right)^{3}\)

=>\(x^{3} - 6 x^{2} + x - 6 < = x^{3} - 6 x^{2} + 12 x - 8\)

=>x-6<=12x-8

=>-11x<=-8+6=-2

=>\(x > = \frac{2}{11}\)

a: \(\frac{3 x + 5}{2} - x > = 1 + \frac{x + 2}{3}\)

=>\(\frac{3 x + 5 - 2 x}{2} > = \frac{3 + x + 2}{3}\)

=>\(\frac{x + 5}{2} - \frac{x + 5}{3} > = 0\)

=>\(\frac{3 \left(\right. x + 5 \left.\right) - 2 \left(\right. x + 5 \left.\right)}{6} > = 0\)

=>\(\frac{x + 5}{6} > = 0\)

=>x+5>=0

=>x>=-5

b: \(\frac{x - 2}{3} - x - 2 < = \frac{x - 17}{2}\)

=>\(\frac{2 \left(\right. x - 2 \left.\right)}{6} + \frac{6 \left(\right. - x - 2 \left.\right)}{6} < = \frac{3 \left(\right. x - 17 \left.\right)}{6}\)

=>\(2 \left(\right. x - 2 \left.\right) + 6 \left(\right. - x - 2 \left.\right) < = 3 \left(\right. x - 17 \left.\right)\)

=>\(2 x - 4 - 6 x - 12 < = 3 x - 51\)

=>-4x-16<=3x-51

=>-7x<=-35

=>x>=5

c: \(\frac{2 x + 1}{3} - \frac{x - 4}{4} < = \frac{3 x + 1}{6} - \frac{x - 4}{12}\)

=>\(\frac{4 \left(\right. 2 x + 1 \left.\right) - 3 \left(\right. x - 4 \left.\right)}{12} < = \frac{2 \left(\right. 3 x + 1 \left.\right) - x + 4}{12}\)

=>4(2x+1)-3(x-4)<=2(3x+1)-x+4

=>8x+4-3x+12<=6x+2-x+4

=>5x+16<=5x+6

=>16<=6(sai)

Vậy: BPT vô nghiệm

a: \(\frac{3 \left(\right. 2 x + 1 \left.\right)}{20} + 1 > \frac{3 x + 52}{10}\)

=>\(\frac{6 x + 3}{20} + \frac{20}{20} > \frac{6 x + 104}{20}\)

=>6x+23>6x+104

=>23>104(sai)

vậy: \(x \in \emptyset\)

b: \(\frac{4 x - 1}{2} + \frac{6 x - 19}{6} < = \frac{9 x - 11}{3}\)

=>\(\frac{3 \left(\right. 4 x - 1 \left.\right) + 6 x - 19}{6} < = \frac{2 \left(\right. 9 x - 11 \left.\right)}{6}\)

=>12x-3+6x-19<=18x-22

=>-22<=-22(luôn đúng)

Vậy: \(x \in R\)

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.

Vẽ \(A K ⊥ B C\) tại K, \(A H ⊥ \&\text{nbsp}; D C\) tại \(H\).

Khi đó tứ giác \(A K C H\) là hình chữ nhật nên \(A K = C H\); \(A H = C K\)

Trong tam giác vuông \(A K B\) vuông tại \(K\) có \(A B = 10\) cm, \(\hat{A B K} = 7 0^{\circ}\) 

\(A K = A B . \&\text{nbsp}; sin ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\) suy ra \(A K = C H = 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(D H = C D - H C = 15 - 10. sin ⁡ \&\text{nbsp}; 7 0^{\circ}\)

\(B K = A B . cos ⁡ \&\text{nbsp}; 7 0^{\circ} = 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Suy ra \(C K = C B - B K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

hay \(A H = C K = 13 - 10. cos ⁡ \&\text{nbsp}; 7 0^{\circ}\)

Theo định lí Pythagore trong tam giác vuông \(A D H\):

\(A D = \sqrt{A H^{2} + D H^{2}} = \sqrt{\left(\right. 13 - 10. cos ⁡ 7 0^{\circ} \left.\right)^{2} + \left(\right. 15 - 10. sin ⁡ 7 0^{\circ} \&\text{nbsp}; \left.\right)^{2}} \approx 11 , 1\) m.