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\(\frac{x-2}{27}\)+\(\frac{x-3}{26}\)+\(\frac{x-4}{25}\)+\(\frac{x-44}{5}\)=1
<=> \(\frac{x-2}{27}-1\)+\(\frac{x-3}{26}-1\)+\(\frac{x-4}{25}-1\)+\(\frac{x-44}{5}+3\)=1
<=> \(\frac{x-29}{27}\)+\(\frac{x-29}{26}\)+\(\frac{x-29}{25}\)+\(\frac{x-29}{5}\)=1
<=> ( x- 29 ) \(\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\right)\)=1
phần sau tự làm tp nhé!
(\(\frac45:\frac65\) + \(\frac15\) : \(\frac{1}{x}\)) x 30 - 26 = 54
(\(\frac45\times\frac56\) + \(\frac15\times x\)) x 30 = 54 + 26
(\(\frac23+\frac15x\)) x 30 = 80
\(\frac23+\frac15x\) = 80 : 30
\(\frac23+\frac15x=\frac83\)
\(\frac15x=\frac83-\frac23\)
\(\frac15x\) = 2
\(x=2:\frac15\)
\(x=2\times5\)
\(x=10\)
Vậy \(x=10\)
6,3 x y + 3,7 x y = 100
y x (6,3 + 3,7) = 100
y x 10 = 100
y = 100 : 10
y = 10
Vậy y = 10
a, 15 - 2 | x + \(\dfrac{1}{3}\)| = \(\dfrac{4}{7}\)
=>2| x + \(\dfrac{1}{3}\)| = 15 - \(\dfrac{4}{7}\)
=> 2 | x + \(\dfrac{1}{3}\)| =\(\dfrac{101}{7}\)
=> | x + \(\dfrac{1}{3}\)| = \(\dfrac{101}{14}\)
=> x+ \(\dfrac{1}{3}\)= \(\dfrac{101}{14}\) hoặc x +\(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)
+) x+\(\dfrac{1}{3}\)= \(\dfrac{101}{14}\)=> x = \(\dfrac{302}{3}\)
+) x + \(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)=> x =-\(\dfrac{317}{42}\)
Vậy ...........
b, \(\dfrac{13}{11}\).\(\dfrac{22}{26}\)-x2=\(\dfrac{7}{16}\)
=>1 -x2 = \(\dfrac{7}{16}\)
=> x2= \(\dfrac{9}{16}\)
=> x= \(\dfrac{3}{4}\)hoặc x = -\(\dfrac{3}{4}\)
Vậy ..................
\(a,15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)
\(2\left|x+\dfrac{1}{3}\right|=15-\dfrac{4}{7}\)
\(2\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}\)
\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}:2\)
\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{14}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{101}{14}\) hoặc \(x+\dfrac{1}{3}=-\dfrac{101}{14}\)
\(x=\dfrac{101}{14}-\dfrac{1}{3}\) \(x=-\dfrac{101}{14}-\dfrac{1}{3}\)
\(x=\dfrac{302}{3}\) \(x=-\dfrac{317}{42}\)
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(=>\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}}\)
Vậy x thuộc {-1/2 ; 1/3}
\(x.3\frac{1}{4}+\frac{-7}{6}.x-1\frac{2}{3}=\frac{5}{12}.2\)
\(x.\frac{13}{4}+\frac{-7}{6}.x-\frac{5}{3}=\frac{5}{6}\)
\(x.\left(\frac{13}{4}+\frac{-7}{6}\right)=\frac{5}{6}+\frac{5}{3}\)
\(x.\left(\frac{39}{12}+\frac{-14}{12}\right)=\frac{5}{6}+\frac{10}{6}\)
\(x.\frac{25}{12}=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{25}{12}\)
\(x=\frac{5}{2}.\frac{12}{25}\)
\(x=\frac{6}{5}\)
a) 100x +101.50 = 5950
100 x = 900
x = 9
b) (1.2 + 1.2.3.4) . x = 26
26 .x = 26
x =1
2) 11a + 11b = 55
a+b =5 = 1+4 = 2+3
ab = 14;41;23;32.
a,Ta có: (x + 1) + (x + 2) + (x + 3) +....+ ( x + 100) = 5950
=> (x + x + x + .... + x) + (1 + 2 + 3 + .... + 100) = 5950
=> 100x + 5050 = 5950
=> 100x = 5950 - 5050
=> 100x = 900
=> x = 900 : 100
=> x = 9
\(3\dfrac{1}{2}+15\dfrac{4}{9}:\dfrac{26}{7}-2\dfrac{4}{9}:\dfrac{26}{7}\)
\(=\dfrac{7}{2}+\dfrac{139}{9}:\dfrac{26}{7}-\dfrac{22}{9}:\dfrac{26}{7}\)
\(=\dfrac{7}{2}+\left(\dfrac{139}{9}-\dfrac{22}{9}\right):\dfrac{26}{7}\)
\(=\dfrac{7}{2}+13:\dfrac{26}{7}\)
\(=\dfrac{7}{2}+\dfrac{7}{2}\)
\(=7.\)
\(3\dfrac{1}{2}+15\dfrac{4}{9}:\dfrac{26}{7}-2\dfrac{4}{9}:\dfrac{26}{7}\)
=\(\dfrac{7}{2}+\dfrac{139}{9}:\dfrac{26}{7}-\dfrac{22}{9}:\dfrac{26}{7}\)
=\(\dfrac{7}{2}+\left(\dfrac{139}{9}-\dfrac{22}{9}\right):\dfrac{26}{7}\)
=\(\dfrac{7}{2}+13:\dfrac{26}{7}\)
=\(\dfrac{7}{2}+13.\dfrac{7}{26}\)
=\(\dfrac{7}{2}+\dfrac{7}{2}\)
= 7
Bài 1:
\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
......????
(x-26).2= 16
x-26 = 16:2
x-26 = 8
x = 8+26
x = 34
Vậy x=34
giup to voi
\(\left(x-26\right).2=4^2\)
\(\left(x-26\right).2=8\)
\(\left(x-26\right)=8:2\)
\(\) \(x-26=4\)
\(x=4+26\)
\(x=30\) \(\)
@ Nguyễn Thị Sen làm sai r 4^2 =16 mà bạn
(x - 26) . 2 = 4\(^2\)
(x - 26) . 2 = 16
x - 26 = 16 : 2
x - 26 = 8
x = 8 + 26
x = 34
Vậy x = 34
@ Nguyễn thị Sen, 4^2 là 4 . 4 chứ không phải 4 . 2 đâu nha bạn.