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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 9
Kết luận của định lý ứng với hình vẽ là:
\(\hat{tOz}\) = 90\(^0\)
11 tháng 9
Bài 2:
a: \(A=\frac17+\frac{1}{7^2}+\cdots+\frac{1}{7^{100}}\)
=>\(7A=1+\frac17+\cdots+\frac{1}{7^{99}}\)
=>\(7A-A=1+\frac17+\cdots+\frac{1}{7^{99}}-\frac17-\frac{1}{7^2}-\cdots-\frac{1}{7^{100}}\)
=>\(6A=1-\frac{1}{7^{100}}=\frac{7^{100}-1}{7^{100}}\)
=>\(A=\frac{7^{100}-1}{6\cdot7^{100}}\)
b: \(B=\frac53+\frac{5}{3^2}+\frac{5}{3^3}+\cdots+\frac{5}{3^{20}}\)
=>\(3B=5+\frac53+\frac{5}{3^2}+\cdots+\frac{5}{3^{19}}\)
=>\(3B-B=5+\frac53+\frac{5}{3^2}+\cdots+\frac{5}{3^{19}}-\frac53-\frac{5}{3^2}-\cdots-\frac{5}{3^{20}}\)
=>\(2B=5-\frac{5}{3^{20}}=\frac{5\cdot3^{20}-5}{3^{20}}\)
=>\(B=\frac{5\cdot3^{20}-5}{2\cdot3^{20}}\)
c: \(C=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\cdots+\frac{1}{3^{50}}\)
=>\(3C=-1+\frac13-\frac{1}{3^2}+\frac{1}{3^3}-\cdots+\frac{1}{3^{49}}\)
=>\(3C+C=-1+\frac13-\frac{1}{3^2}+\frac{1}{3^3}-\cdots+\frac{1}{3^{49}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\cdots+\frac{1}{3^{50}}\)
=>\(4C=-1+\frac{1}{3^{50}}=\frac{-3^{50}+1}{3^{50}}\)
=>\(C=\frac{-3^{50}+1}{4\cdot3^{50}}\)
d: \(D=\left(-\frac17\right)^0+\left(-\frac17\right)^1+\left(-\frac17\right)^2+\cdots+\left(-\frac17\right)^{2017}\)
=>\(D=1-\frac17+\frac{1}{7^2}-\frac{1}{7^3}+\cdots-\frac{1}{7^{2017}}\)
=>\(7D=7-1+\frac17-\frac{1}{7^2}+\cdots-\frac{1}{7^{2016}}\)
=>\(7D+D=7-1+\frac17-\frac{1}{7^2}+\cdots-\frac{1}{7^{2016}}+1-\frac17+\frac{1}{7^2}-\frac{1}{7^3}+\cdots-\frac{1}{7^{2017}}\)
=>\(8D=7-\frac{1}{7^{2017}}=\frac{7^{2018}-1}{7^{2017}}\)
=>\(D=\frac{7^{2018}-1}{8\cdot7^{2017}}\)
e: \(E=\frac12+\frac{1}{2^3}+\frac{1}{2^5}+\cdots+\frac{1}{2^{99}}\)
=>\(4E=2+\frac12+\frac{1}{2^3}+\cdots+\frac{1}{2^{97}}\)
=>\(4E-E=2+\frac12+\frac{1}{2^3}+\cdots+\frac{1}{2^{97}}-\frac12-\frac{1}{2^3}-\frac{1}{2^5}-\cdots-\frac{1}{2^{99}}\)
=>\(3E=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\)
=>\(E=\frac{2^{100}-1}{3\cdot2^{99}}\)
Bài 1:
a: \(A=2\cdot4+4\cdot6+6\cdot8+\cdots+98\cdot100\)
\(=4\left(1\cdot2+2\cdot3+3\cdot4+\cdots+49\cdot50\right)\)
\(=4\left\lbrack1\left(1+1\right)+2\left(2+1\right)+3\left(3+1\right)+\cdots+49\left(49+1\right)\right\rbrack\)
\(=4\left\lbrack\left(1^2+2^2+\cdots+49^2\right)+\left(1+2+3+\cdots+49\right)\right\rbrack\)
\(=4\cdot\left\lbrack\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}+\frac{49\cdot50}{2}\right\rbrack=4\cdot\left\lbrack\frac{49\cdot50\cdot99}{6}+49\cdot25\right\rbrack\)
\(=4\cdot\left\lbrack49\cdot25\cdot33+49\cdot25\right\rbrack=4\cdot49\cdot25\cdot34=100\cdot49\cdot34\)
=166600
b: \(B=1\cdot99+2\cdot98+\cdots+97\cdot3+98\cdot2+99\cdot1\)
\(=2\cdot\left(1\cdot99+2\cdot98+\cdots+48\cdot52+49\cdot51\right)+50^2\)
\(=2\cdot\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+48\left(100-48\right)+49\left(100-49\right)\right\rbrack+50^2\)
\(=2\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\) +2500
\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot49\cdot25-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot49\cdot25-49\cdot25\cdot33\right\rbrack+2500=2\cdot25\cdot49\left(100-33\right)+2500\)
\(=50\cdot49\cdot67+2500=166650\)
d: \(D=2^2+4^2+\cdots+98^2+100^2\)
\(=2^2\left(1^2+2^2+\cdots+49^2+50^2\right)\)
\(=4\cdot\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}=4\cdot\frac{50\cdot51\cdot101}{6}\)
\(=4\cdot25\cdot17\cdot101=100\cdot17\cdot101=171700\)
e: \(E=1^2+3^2+5^2+\cdots+99^2\)
\(=\left(1^2+2^2+3^2+4^2+\cdots+99^2+100^2\right)-\left(2^2+4^2+\cdots+100^2\right)\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}-2^2\left(1^2+2^2+\cdots+50^2\right)\)
\(=\frac{100\cdot101\cdot201}{6}-4\cdot\frac{50\left(50+1\right)\left(2\cdot50+1\right)}{6}\)
\(=50\cdot101\cdot67-4\cdot\frac{50\cdot51\cdot101}{6}\)
\(=50\cdot101\cdot67-4\cdot25\cdot17\cdot101=101\cdot50\left(67-2\cdot17\right)\)
\(=50\cdot101\cdot33=166650\)
f: \(F=1^2-2^2+3^2-4^2+\cdots+99^2-100^2\)
\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+\cdots+\left(99-100\right)\left(99+100\right)\)
=-(1+2+3+4+...+99+100)
\(=-100\cdot\frac{101}{2}=-50\cdot101=-5050\)
3 tháng 9
a: ta có: \(\hat{xAB}+\hat{yBA}=45^0+135^0=180^0\)
mà hai góc này là hai góc ở vị trí trong cùng phía
nên Ax//By
b: Gọi BM là tia đối của tia By
Khi đó, ta có: \(\hat{MBA}+\hat{yBA}=180^0\) (hai góc kề bù)
=>\(\hat{MBA}=180^0-135^0=45^0\)
Ta có: tia BM nằm giữa hai tia BA và BC
=>\(\hat{ABM}+\hat{CBM}=\hat{ABC}\)
=>\(\hat{CBM}=75^0-45^0=30^0\)
Ta có: \(\hat{MBC}=\hat{BCz}\left(=30^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên By//Cz
3 tháng 9
a: ta có: \(\hat{tKy}+\hat{tKm}=180^0\) (hai góc kề bù)
=>\(\hat{tKm}=180^0-150^0=30^0\)
Ta có: \(\hat{tNz}=\hat{tKm}\left(=30^0\right)\)
mà hai góc này là hai góc ở vị trí đồng vị
nên Nz//Km
b: Ta có: \(\hat{tKy}+\hat{tKM}+\hat{yKM}=360^0\)
=>\(\hat{yKM}=360^0-90^0-150^0=120^0\)
Ta có: \(\hat{yKM}=\hat{KMn}\left(=120^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên Ky//Mn
d: \(\frac27-\left(\frac23+2x\right)=\frac57\)
=>\(2x+\frac23=\frac27-\frac57=-\frac37\)
=>\(2x=-\frac37-\frac23=-\frac{9}{21}-\frac{14}{21}=-\frac{23}{21}\)
=>\(x=-\frac{23}{21}:2=-\frac{23}{42}\)
e: \(\frac12-2x=\left(-\frac12\right)^3\)
=>\(\frac12-2x=-\frac18\)
=>\(2x=\frac12+\frac18=\frac58\)
=>\(x=\frac58:2=\frac{5}{16}\)
f: \(\left(2x-3\right)\left(\frac34x+1\right)=0\)
=>\(\left[\begin{array}{l}2x-3=0\\ \frac34x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3\\ \frac34x=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\\ x=-\frac43\end{array}\right.\)
g: \(\frac{7}{12}-\left(x+\frac76\right):\frac65=-\frac54\)
=>\(\left(x+\frac76\right):\frac65=\frac{7}{12}+\frac54=\frac{7}{12}+\frac{15}{12}=\frac{22}{12}=\frac{11}{6}\)
=>\(x+\frac76=\frac{11}{6}\cdot\frac65=\frac{11}{5}\)
=>\(x=\frac{11}{5}-\frac76=\frac{66}{30}-\frac{35}{30}=\frac{31}{30}\)
h: \(\frac34:\left(x+\frac12\right)-\frac56=-\frac14\)
=>\(\frac34:\left(x+\frac12\right)=-\frac14+\frac56=-\frac{3}{12}+\frac{10}{12}=\frac{7}{12}\)
=>\(x+\frac12=\frac34:\frac{7}{12}=\frac34\cdot\frac{12}{7}=\frac{36}{28}=\frac97\)
=>\(x=\frac97-\frac12=\frac{18}{14}-\frac{7}{14}=\frac{11}{14}\)
i: \(\frac25x+\frac35x=\frac34\)
=>\(x\left(\frac25+\frac35\right)=\frac34\)
=>\(x\cdot\frac55=\frac34\)
=>\(x=\frac34\)
k: \(\frac12x+\frac23x-x=\frac13\)
=>\(x\left(\frac12+\frac23-1\right)=\frac13\)
=>\(x\left(\frac12-\frac13\right)=\frac13\)
=>\(x\cdot\frac16=\frac13\)
=>\(x=\frac13:\frac16=2\)
l: \(\left(\frac32-\frac{2}{-5}\right):x-\frac12=\frac32\)
=>\(\left(\frac32+\frac25\right):x=\frac32+\frac12=2\)
=>\(\left(\frac{15}{10}+\frac{4}{10}\right):x=2\)
=>\(\frac{19}{10}:x=2\)
=>\(x=\frac{19}{10}:2=\frac{19}{20}\)
m: \(\left(5x-1\right)\left(2x-\frac13\right)=0\)
=>\(\left[\begin{array}{l}5x-1=0\\ 2x-\frac13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=1\\ 2x=\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac15\\ x=\frac16\end{array}\right.\)
khó