\(N=-1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

Xét \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

\(\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{11}}\Rightarrow\dfrac{1}{2}A-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{11}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(\Leftrightarrow-\dfrac{1}{2}A=-\dfrac{1}{2}+\dfrac{1}{2^{11}}\Rightarrow A=-\dfrac{1}{2^{10}}\)

\(\Rightarrow N=-1-\left(-\dfrac{1}{2^{10}}\right)=-1+\dfrac{1}{2^{10}}\) 

=> Vậy ko tm đpcm 

\(\dfrac{x}{2}+\dfrac{x}{3}-1=\dfrac{1}{6}\Rightarrow3x+2x-6=1\Leftrightarrow5x=7\Leftrightarrow x=\dfrac{7}{5}\)

a) Do ∆ABC cân tại A (gt)

⇒ AB = AC và ∠ABC = ∠ACB

Ta có:

∠ABF + ∠ABC = 180⁰ (kề bù)

∠ACE + ∠ACB = 180⁰ (kề bù)

Mà ∠ABC = ∠ACB (cmt)

⇒ ∠ABF = ∠ACE

Xét ∆ABF và ∆ACE có:

AB = AC (cmt)

∠ABE = ∠ACF (cmt)

BF = CE (gt)

⇒ ∆ABF = ∆ACE (c-g-c)

⇒ AF = AE (hai cạnh tương ứng)

⇒ ∆AEF cân tại A

b) *) Cách 1:

Do ∆ABF = ∆ACE (cmt)

⇒ ∠BAF = ∠CAE (hai góc tương ứng)

⇒ ∠BAH = ∠CAK

Xét hai tam giác vuông: ∆ABH và ∆ACK có:

AB = AC (cmt)

∠BAH = ∠CAK (cmt)

⇒ ∆ABH = ∆ACK (cạnh huyền - góc nhọn)

⇒ BH = CK (hai cạnh tương ứng)

*) Cách 2:

Do ∆AEF cân tại A (cmt)

⇒ ∠AFE = ∠AEF

⇒ ∠HFB = ∠KEC

Xét hai tam giác vuông: ∆BHF và ∆CKE có:

BF = CE (gt)

∠HFB = ∠KEC (cmt)

⇒ ∆BHF = ∆CKE (cạnh huyền - góc nhọn)

⇒ BH = CK (hai cạnh tương ứng)

c) Sửa đề: Gọi O là giao điểm của HB và KC

Do ∆BHF = ∆CKE (cmt)

⇒ ∠HBF = ∠KCE (hai góc tương ứng)

Mà ∠CBO = ∠HBF (đối đỉnh)

∠BCO = ∠KCE (đối đỉnh)

⇒ ∠CBO = ∠BCO

⇒ ∆BOC cân tại O

Sửa đề: \(B=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2022\cdot2024}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\cdot\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2023^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2023^2}{2023^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2023}{1\cdot2\cdot...\cdot2022}\cdot\dfrac{2\cdot3\cdot....\cdot2023}{3\cdot4\cdot...\cdot2024}\)

\(=\dfrac{2023}{1}\cdot\dfrac{2}{2024}=\dfrac{2023}{1012}\)

\(\widehat{zOn}-\dfrac{1}{2}\cdot\widehat{yOn}\)

\(=90^0-\widehat{zOm}-\dfrac{1}{2}\left(180^0-\widehat{xOn}\right)\)

\(=90^0-\widehat{zOm}-90^0+\dfrac{1}{2}\cdot\widehat{xOn}\)

\(=\dfrac{1}{2}\cdot\widehat{xOn}-\widehat{zOm}=\dfrac{1}{2}\left(\widehat{xOn}-\widehat{xOm}\right)=\dfrac{1}{2}\cdot\widehat{nOm}=45^0\)

Sửa đề: x-2y+3z=-33

10x=6y=5z

=>\(\dfrac{10x}{30}=\dfrac{6y}{30}=\dfrac{5z}{30}\)

=>\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}\)

mà x-2y+3z=-33

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-2y+3z}{3-2\cdot5+3\cdot6}=\dfrac{-33}{11}=-3\)

=>\(x=-3\cdot3=-9;y=-3\cdot5=-15;z=-3\cdot6=-18\)

\(\left(\dfrac{1}{6}+\dfrac{1}{3}\right)^2:\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)\)

\(=\left(\dfrac{1}{6}+\dfrac{2}{6}\right)^2:\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{15}{12}\right)\)

\(=\left(\dfrac{3}{6}\right)^2:\dfrac{5}{12}=\dfrac{1}{4}\cdot\dfrac{12}{5}=\dfrac{3}{5}\)

Vì \(\widehat{A_1}\ne\widehat{B_2}\)

nên b sẽ không song song với c

mà b và c là hai đường thẳng phân biệt

nên b cắt c

\(C=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{301\cdot304}\\ =2\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{301\cdot304}\right)\\ =2\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{301}-\dfrac{1}{304}\right)\\ =2\cdot\left(1-\dfrac{1}{304}\right)\\ =2\cdot\dfrac{303}{304}\\ =\dfrac{303}{152}\) 

\(B=\dfrac{11}{210}-\left(\dfrac{16}{15\cdot31}+\dfrac{13}{31\cdot44}+\dfrac{16}{44\cdot60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\dfrac{1}{20}\\ =\dfrac{1}{420}\)

\(a.\left(\dfrac{-1}{2}\right)^2\cdot\left(\dfrac{2}{5}\right)^2 \\ =\left(\dfrac{-1}{2}\cdot\dfrac{2}{5}\right)^2\\ =\left(\dfrac{-1}{5}\right)^2\\ =\dfrac{1}{25}\\ b.\left(\dfrac{1}{9}\right)^2:\left(\dfrac{1}{3}\right)^3\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^2:\left(\dfrac{1}{3}\right)^3\\ =\left(\dfrac{1}{3}\right)^4:\left(\dfrac{1}{3}\right)^3\\ =\dfrac{1}{3}\\ c.\left(\dfrac{-1}{2}\right)^3\cdot\left(\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-1}{2}\cdot\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-3}{4}\right)^3\\ =\dfrac{-27}{64}\)