câu 4,5 sai r bn

\(1,\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}}=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)

\(2,\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

\(3,\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=\sqrt{3}-\sqrt{3}=0\)

\(4,\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{\sqrt{3}-1}=\sqrt{2}-\sqrt{2}=0\)

\(5,\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)

\(6,\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}+\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}+\sqrt{5}}{2}=\dfrac{3\sqrt{5}}{2}\)

Phần a sai đề nha phải là CM C chia hết cho 13

\(a,C=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+3^9\left(1+3+3^2\right)\\ C=\left(1+3+3^2\right)\left(1+3^3+3^6+3^9\right)\\ C=13\left(1+3^3+3^6+3^9\right)⋮13\)

\(b,C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(+3^8+3^9+3^{10}+3^{11}\right)\\ C=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\\ C=\left(1+3+3^2+3^3\right)\left(1+3+3^8\right)\\ C=40\left(1+3^4+3^8\right)⋮40\)

\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)

\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)

\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)

\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)

\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)

AH=tan30\(\times\)BH

AH=tan40\(\times\)(250-BH)

\(\Rightarrow\)tan30.BH=tan40(250-BH)

\(\Rightarrow\)BH\(\approx\)148(m)

\(\Rightarrow\)d\(\approx\)85,4

16B

17B

18A

19A

20B

21D

22A

23B

24A

25A

26B

27D

28C

3.2+3.4+....+3.200

=3(2+4+6+.....+200)

=3((2+200)+(4+198)+...+(100+102))

=3(202\(\times\)50)=30300

a)Vì cửa hàng giảm 20% cho tất cả sản phẩm nên giá chiếc váy là:

\(800000-800000\times20\%=640000\left(dong\right)\)

Mà chị thanh là khách hàng thân thiết nên chị được giảm thêm 10%, giá chiếc váy còn:

\(640000-640000\times10\%=576000\left(đong\right)\)

vậy giá chiếc váy là 576000(đồng)

b)Giá chiếc túi sách khi cô chưa được giảm 10% là:

\(864000\times110\%=950400\)

Giá chiếc váy lúc ban đầu là:

\(950400\times120\%=1150480\)

Vậy.....