tìm x biết\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+3}{250}+\frac{x+3}{240}\)
tìm x biết
\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+30}{250}+\frac{x+40}{240}\)
HD dùng PP Quy đồng tử (không quy đồng Mẫu)
\(\left(\frac{x+10}{270}+10\right)+\left(\frac{x+20}{260}+10\right)=\left(\frac{x+30}{250}+10\right)+\left(\frac{x+40}{240}+10\right)\\ \)
\(\left(x+280\right)\left(....\right)=0\)chú ý (...) thường khác không nếu bằng =0=> đúng với mọi x
nếu khác không=> x=-280
tìm x giúp m nha đúng mình tíck
\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+3}{250}+\frac{x+3}{240}\)
\(\Leftrightarrow\frac{x+10}{270}+1+\frac{x+20}{260}+1=\frac{x+30}{250}+1+\frac{x+40}{240}+1\)
\(\Leftrightarrow\frac{x+280}{270}+\frac{x+280}{260}=\frac{x+280}{250}+\frac{x+280}{240}\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}\right)=\left(x+280\right)\left(\frac{1}{250}+\frac{1}{240}\right)\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}-\frac{1}{250}-\frac{1}{240}\right)=0\)
=>x=-280
Tìm x biết :
a) \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{50}\)
b) \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)
c) \(\frac{x-10}{20}+\frac{x-20}{10}+\frac{x-30}{5}=\frac{x-14}{4}\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\) hả đề này mk làm đc
\(\left(3\cdot x-0,8\right):x+14,5=15\)
1,2\(\cdot\)(\(\frac{2,4\cdot x-0,23}{x}\)\(-0,05=1,44\)
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}=\frac{0,33\cdot x}{2009}\)
\(x\)-\(\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\left(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\right)\)\(\cdot x=\frac{5}{14}\)
Các bạn ơi trả lời giùm mình với nhé, cần gấp.
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
a) Tính A=\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+..........+\frac{10}{1400}\)
b) Tìm x thuộc Z , biết \(\frac{1.2+2.3+3.4+....+99.100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+.....+\left(x^2+99\right)}=50\frac{116}{131}\)
TÌM x BIẾT:
a,\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
với x\(\notin\){-2;-5;-10;-17}
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
với x\(\notin\){1;3;8;20}
c, TÌM X BIẾT:
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
GIÚP MÌNH CHÚT NHA MÌNH CẦN NGAY. THANKS!
Tìm x, biết:
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\left(x\notin-2;-5;-10;-17\right)\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
Với \(x\notin1;3;8;20\)
\(\frac{x+1}{10}+\frac{2+1}{11}\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
Tìm X biết : X x \(\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
3 + \(\frac{3}{20}\)+ \(\frac{3}{13}\) + \(\frac{3}{2013}\)
X x = \(\frac{5}{3}\)
5 + \(\frac{5}{20}\) + \(\frac{5}{13}\) + \(\frac{5}{2013}\)
3 x ( 1 + \(\frac{1}{20}\) + \(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x = \(\frac{5}{3}\)
5 x ( 1 + \(\frac{1}{20}\) +\(\frac{1}{13}\) + \(\frac{1}{2013}\) )
X x \(\frac{3}{5}\) = \(\frac{5}{3}\) => X = \(\frac{25}{9}\) vậy X = \(\frac{25}{9}\)
Ta có : \(X.\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{5}{3}\)
\(\Leftrightarrow X.\frac{3}{5}=\frac{5}{3}\Rightarrow X=\frac{5}{3}:\frac{3}{5}=\frac{5}{3}.\frac{5}{3}=\frac{25}{9}\)
CAU 1)\(\frac{X+Y}{z}=\frac{y+z}{x}=\frac{x+z}{y}\)
Khi đó x+y=kz Khi đó K=
CAU 2)Tìm X biết :
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}+\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
X =
CAU 3)tÍNH
\(S=\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{20}+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{21}{20}=\)