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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(\frac{9}{10}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ ........ + \(\frac{1}{90}\)
= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ ....... + \(\frac{1}{9.10}\)
= \(\frac{2-1}{1.2}\)+ \(\frac{3-2}{2.3}\)+ \(\frac{4-3}{3.4}\)+ \(\frac{5-4}{4.5}\)+ \(\frac{6-5}{5.6}\)+ ......... + \(\frac{10-9}{9.10}\)
= \(\frac{2}{1.2}\)- \(\frac{1}{1.2}\)+ \(\frac{3}{2.3}\)- \(\frac{2}{2.3}\)+ \(\frac{4}{3.4}\)- \(\frac{3}{3.4}\)+ \(\frac{5}{4.5}\)- \(\frac{4}{4.5}\)+ \(\frac{6}{5.6}\)- \(\frac{5}{5.6}\)+ ........ + \(\frac{10}{9.10}\)- \(\frac{9}{9.10}\)
= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{6}\)+ ........... + \(\frac{1}{9}\)- \(\frac{1}{10}\)
Sau đó ta trực tiêu:
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta được:
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=3\left(x+6\right)\)
\(\Leftrightarrow x^2+3x+2=3x+18\)
\(\Leftrightarrow x^2=16\)
Vậy \(x\in\left\{4;-4\right\}\)
(x+2)/(x+6)=3/(x+1)
<=> (x+2)(x+1)/(x+6)(x+1)=3(x+6)/(x+6)(x+1)
=>(x+2)(x+1)=3(x+6)
<=> x^2+x+2x+2=3x+18
<=> x^2=16
<=>x^2=4^2 hoặc (-4)^2
<=> x=4 hoặc x=-4
Vậy.........
A = 1/2 + 1/6 + 1/12 + .. + 1/6480
A= 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/80 x 81
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/80 - 1/81
A = 1 - 1/81
A = 80/81
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)
\(A=\frac{1}{1}\times2+\frac{1}{2}\times3+\frac{1}{3}\times4+...+\frac{1}{80}\times81\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{80}-\frac{1}{81}\)
\(A=1-\frac{1}{81}\)
\(A=\frac{80}{81}\)
\(\frac{y}{12}=\frac{x}{4}=\frac{y-x}{12-4}=\frac{4}{8}=\frac{1}{2}.\)
Từ đó tính được x và y => Z
Áp dụng tính chất của dãy tỉ số bằng nhau ta được :
\(\frac{x}{4}=\frac{y}{12}=\frac{y-x}{12-4}=\frac{4}{8}=\frac{1}{2}\)
Do đó : \(\hept{\begin{cases}\frac{x}{4}=\frac{1}{2}\\\frac{y}{12}=\frac{1}{2}\\\frac{z}{15}=\frac{1}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=6\\z=7,5\end{cases}}\)
Vậy .........
\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=7-5x\)
\(\Rightarrow x^2-2x-3x+6=7-5x\)
\(\Rightarrow x^2-2x-3x+5x=7-6\)
\(\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
k nhé,Vy Nguyễn Đặng Khánh !
nhân tích chéo
\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=1\left(7-5x\right)\)
\(\Leftrightarrow x^2-3x-2x+6=7-5x\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow x^2=1\Leftrightarrow x=1\)
vậy x=1
Có \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{90}=\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{9.10}\)
= \(1-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{9}-\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
kết quả là \(\frac{1}{10}\)nhà mình không biết cách làm vậy cho mình xin lỗi nha!
mình trả lời đầu tiên đó nha!
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{10}{9}\cdot\frac{11}{10}\)
\(=\frac{3.4.5.....10.11}{2.3.4....10}=\frac{11}{2}\)
\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)
\(Z=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)
\(Z=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)
\(Z=\frac{1}{1}-\frac{1}{81}=\frac{80}{81}\)
\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{6480}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{80.81}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....+\frac{1}{80}-\frac{1}{81}\)
\(=1-\frac{1}{81}\)
\(=\frac{80}{81}\)