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a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150
=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150
=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37
=> [497 - (8 x X +11): X ] : 3 - 78 = 37
=> [497 - (8 x X + 11) : X] : 3 = 115
=> 497 - ( 8 x X + 11) : X = 345
=> (8 x X + 11) : X = 497 - 345 = 152
=> 8X + 11 = 152X
=> 152X - 8X = 11
=> 144X = 11
=> X = 11/144
b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)
=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15
=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1
=> \(x\cdot4-\frac{1}{4}=24,15\)
=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)
=> x = 24,4 : 4 = 6,1
Còn câu c tương tự
*19,96+4,19-24,15:(1/4-1/4) *252/x=84/97
=19,96+4,19-24,15:0 x=252x97:84
=19,96+4,19-0 x=291
=24,15-0 * x-2/255=114/153
=24,15 x-2/255=38/51
x=38/51+2/255
x=64/85
a ) ko rõ đề
b )\(\frac{252}{x}=\frac{84}{97}\)
=> 252 . 97 = x . 84
=> x . 84 = 24 444
x = 24 444 : 84
x = 291
Vậy x = 291
c ) x - \(\frac{2}{255}=\frac{114}{153}\)
x = 114/153 + 2/255
x = 64/85
Vậy x = 64/85
\(a,\frac{4-x}{-2}=\frac{8}{x-4}\left(x\ne4\right)\)
\(\Leftrightarrow\frac{x-4}{2}=\frac{8}{x-4}\)
\(\Leftrightarrow\left(x-4\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=16\\x-4=-16\end{cases}\Leftrightarrow\orbr{\begin{cases}x=20\\x=-12\end{cases}}}\)(tmđk)
b) \(\frac{3}{x-1}=\frac{x}{24}\left(x\ne1\right)\)
<=> x(x-1)=72
Vì x và x-1 là 2 số liên tiếp => x(x-1)=8.9
=> x=8 (tmđk)
a, ĐKXĐ \(x-4\ne0\Leftrightarrow x\ne4\)
\(\frac{4-x}{-2}=\frac{8}{x-4}\Leftrightarrow-\frac{4-x}{2}=\frac{8}{x-4}\Leftrightarrow x-4=\frac{16}{x-4}\Leftrightarrow\left(x-4\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=\sqrt{16}=4\\x-4=\sqrt{16}=-4\end{cases}}\)Theo ĐKXĐ => x = -4
b, ĐKXĐ \(x-1\ne0\Leftrightarrow x\ne1\)
\(\frac{3}{x-1}=\frac{x}{24}\Leftrightarrow72=x\left(x-1\right)\Leftrightarrow72=x^2-x\Leftrightarrow x^2-x-72=0\Leftrightarrow\left(x-9\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-8\end{cases}}\)Theo ĐKXĐ : x (tm)
\(\frac{3}{4}+\frac{1}{4}:x=\frac{1}{2}\)
\(\frac{1}{4}:x=\frac{1}{2}-\frac{3}{4}\)
\(\frac{1}{4}:x=-\frac{1}{4}\)
\(x=\frac{1}{4}:\left(-\frac{1}{4}\right)\)
\(x=-1\)
~Học tốt~
\(\frac{3}{4}+\frac{1}{4}:x=\frac{1}{2}\)
\(\frac{1}{4}:x=\frac{1}{2}-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-1}{4}\)
\(x=\frac{1}{4}:\frac{-1}{4}\)
\(x=-1\)
\(\frac{1}{7}\cdot\frac{2}{9}+\frac{1}{9}\cdot\frac{3}{7}+\frac{1}{7}\cdot\frac{4}{9}\)
\(=\frac{2}{7}\cdot\frac{1}{9}+\frac{1}{9}\cdot\frac{3}{7}+\frac{4}{7}\cdot\frac{1}{9}\)
\(=\frac{1}{9}\left(\frac{2}{7}+\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{1}{9}\cdot\frac{9}{7}=\frac{1}{7}\)
\(\frac{1}{7}.\frac{2}{9}+\frac{1}{9}.\frac{3}{7}+\frac{1}{7}.\frac{4}{9}\)
\(=\frac{2}{7}.\frac{1}{9}+\frac{1}{9}.\frac{3}{7}+\frac{4}{7}.\frac{1}{9}\)
\(=\frac{1}{9}.\left(\frac{2}{7}+\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{1}{9}.\frac{9}{7}\)
\(=\frac{1}{7}\)
a) \(x+\)\(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(\Rightarrow x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}=-\frac{4}{5}\)
\(\Rightarrow x=\frac{-3}{5}\)
b) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(\Rightarrow2A=1-\frac{1}{2005}\)
\(\Rightarrow2A=\frac{2004}{2005}\)
\(\Rightarrow A=\frac{1002}{2005}\)
Tính tổng:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
= \(\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2003}-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\cdot\frac{2004}{2005}\)
= \(\frac{1002}{2005}\)
k nha
\(19,96+4,19-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24,15\)
\(4x=24,4\)
\(x=6,1\)
\(\text{Bài này cx đơn giản thôi!}\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23.15-19.96\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=4.19-3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15:1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15\)
\(x:\frac{1}{4}=24.15+\frac{1}{4}\)
\(x:\frac{1}{4}=24.4\)
\(x=24.4.\frac{1}{4}\)
\(x=6.1\)