\(\frac{25}{26}+\frac{1}{26}\)
Những câu hỏi liên quan
\(\left(\frac{24}{25}+\frac{25}{26}+\frac{26}{17}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}\right):\frac{3}{b}\)
a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27}Leftrightarrowfrac{x-23}{24}+frac{x-23}{25}-frac{x-23}{26}-frac{x-23}{27}0Leftrightarrowleft(x-23right)left(frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}right)0Leftrightarrow x-230( vì frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}ne0)Leftrightarrow x23Vậy nghiệm của pt x23
Đọc tiếp
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\)( vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\))
\(\Leftrightarrow x=23\)
Vậy nghiệm của pt x=23
\(\frac{6}{5}+\frac{25}{19}-\frac{5}{25}.....\frac{2}{1945}-\frac{2}{1975}+\frac{2}{2013}\) / \(\frac{13}{5}+\frac{13}{19}-\frac{13}{25}.......\frac{26}{1945}-\frac{26}{1976}+\frac{26}{2013}\)
( Hai biểu thức này viết như phân số, cách nhau bằng đường kẻ, đoạn 6/5 ở trên, 13/5 ở dưới
\(\left\{x+\frac{1}{5}\right\}^2+\frac{17}{25}=\frac{26}{25}\)
Ta có: \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
<=> \(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}=\frac{9}{25}\)
<=> \(x+\frac{1}{5}=\frac{9}{25}\)và \(x+\frac{1}{5}=-\frac{9}{25}\)
=> x= \(\frac{4}{25}\) và x=\(-\frac{14}{25}\)
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Ta có:
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{1}{5}=\frac{2}{5}\)
Vậy giá trị của x là \(\frac{2}{5}\)
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Xem thêm câu trả lời
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
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(x+1/5)^2+17/25=26/25
(x+1/5)^2=26/25-17/25
(x+1/5)^2=9/25
\(\Rightarrow\) (x+1/5)^2=(3/25)^2 hoặc (+1/5)^2=(-3/25)^2
⇒x+1/5=3/5 hoặcx+1/5=-3/5
TH1:x+1/5=3/5 TH2:x+1/5=-3/5
x=3/5-1/5 x=-3/5-1/5
x=2/5 x=-4/5
Vậy x∈(2/5;-4/5)
\((x+\frac{1}{2})\times(\frac{2}{3}-2x)=0\)
\(\left\{x+\frac{1}{5}\right\}^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)
TH1 : \(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=\frac{-1}{2}\)
TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)
\(2x=0+\frac{2}{3}=\frac{2}{3}\)
\(x=\frac{2}{3}\div2=\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{2};\frac{1}{3}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}^2-\frac{1}{5}^2\)
\(x=\frac{2}{5}\)
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Tìm x, biết:
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{3}{5}^2\)
\(\left(x+\frac{1}{5}\right)=\frac{3}{5}\)
\(x=\frac{3}{5}-\frac{1}{5}\)
\(x=\frac{2}{5}\)
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(x+1/5) . (x+1/5)=26/25-17/25
x+1/5 .x+1/5=9/25
x+1/5.x=9/25-5/25
x+1/5 . x=4/25
x.(1/5+1)=4/25
x. 6/5=4/25
x=4/25:6/5
x=2/15
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So sánh\(\frac{1}{2}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}+\frac{1}{26}\)và 1
So sánh A và B biết:
a) A =\(\frac{2003.2004-1}{2003.2004}\)và B =\(\frac{2004.2005-1}{2004.2005}\)
b) A =\(\frac{10^{25}+1}{10^{26}+1}\) và B = \(\frac{10^{26}+1}{10^{27}+1}\)