Tìm x nguyên biết:
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right).\left(2x+1\right)}=\dfrac{49}{99}\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{9^{1006}-1}{2}\)
Tìm x biết :\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)
Nguyễn Hoàng Minh cho hỏi 2x + 1 - 1 đâu ra v ạ??
a)tìm các cặp số nguyên x;y thỏa mãn (2x-)(x+1)=|y+1|
b)\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+\left|x+\dfrac{1}{97.99}\right|=50x\)
cho A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)hãy so sánh A với \(\dfrac{-1}{2}\)
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
tìm x:
a,\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right).\left(x-1\right)+\dfrac{1}{10}.x=x-\dfrac{9}{10}\)
b,\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right).\left(x-2\right)+x=\dfrac{149}{99}.x-\dfrac{98}{99}\)
Tìm x: \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2\cdot\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=1-\dfrac{1}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Rightarrow2x+1=99\\ \Leftrightarrow2x=98\\ \Leftrightarrow x=49\)
1) giải phương trình :
a) \(\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
b) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
d) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
i) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
k) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(X-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)
= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)
= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)
=> A=\(\dfrac{7}{15}\)
Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)
=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)
=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)
=> x\(\left(-\dfrac{2}{15}\right)=0\)
=> x=0
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)
<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)
<=> \(\dfrac{-2}{15}x=0\)
<=> \(x=0\)
Vậy: \(s=\left\{0\right\}.\)
Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)