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a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
\(\left(x:\dfrac{x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(x:\dfrac{x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)
\(x:\dfrac{x}{7}+1=\dfrac{1}{7}\)
\(x:\dfrac{x}{7}=\dfrac{1}{7}-1\)
\(x:\dfrac{x}{7}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}.\dfrac{x}{7}\)
\(x=\dfrac{-6.x}{49}\)
\(x=49:\left(-6\right)\)
\(x=-\dfrac{49}{6}\)
a) \(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\)
\(\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\)
\(\left(\dfrac{1}{2}-\dfrac{2}{3}\right)x=\dfrac{7}{12}\)
\(\dfrac{-1}{6}x=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}:\dfrac{-1}{6}\)
\(x=\dfrac{-7}{2}\)
b) \(x:4\dfrac{1}{3}=-2,5\)
\(x:\dfrac{13}{3}=\dfrac{-5}{2}\)
\(x=\dfrac{13}{3}.\dfrac{-5}{2}\)
\(x=\dfrac{-65}{6}\)
c) \(5,5x=\dfrac{13}{15}\)
\(\dfrac{11}{2}x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{11}{2}\)
\(x=\dfrac{26}{165}\)
d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\\ x=\left(-6\right):3\\ x=-2\)
câu d mình nhầm bạn nha
d)\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\dfrac{3x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
Ta có : 3x =-6
x = -6:3
x=-2
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu 1:
\(\left(x-3\right)^2=16\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
vậy x\(\in\)\(\left\{7;-1\right\}\)
câu2:a,
\(\Rightarrow\)\(\dfrac{13}{30}\le x< \dfrac{77}{10}\)
mà x\(\in\)Z\(\Rightarrow\)x\(\in\)\(\left\{1;2;3;4;5;6\right\}\)
vậy x\(\in\)\(\left\{1;2;3;4;5;6\right\}\)
b, ta có: \(\dfrac{-12}{21}=\dfrac{-48}{84}\);\(\dfrac{10}{-28}=\dfrac{-10}{28}=\dfrac{-30}{84}\)
mà \(\dfrac{-48}{84}< \dfrac{-30}{84}\)\(\Rightarrow\)\(\dfrac{-12}{21}< \dfrac{10}{-28}\)
bạn ko nhất thiết phải làm dài dòng ở phần so sánh như thế mình có cách làm ngắn gọn hơn
-12/21<-12/-28
Mà -12/-28<10/-28
=> -12/21<10/-28
1. x3 - \(\dfrac{4}{25}\)x = 0
<=> x(x2 - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh
1 ) \(x^3-\dfrac{4}{25}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)
Vậy .............
2 ) \(3^{4x+4}=9^{x+2}\)
\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)
\(\Leftrightarrow4x+4=2x+4\)
\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)
3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )
4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)
\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)
Vậy ......
5) \(2+4+6+...+2x=210\)
\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)
\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)
\(\Leftrightarrow1+2+3+...+x=105\)
\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)
\(\Leftrightarrow x\left(x+1\right)=210\)
Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)
\(\Leftrightarrow x=14\)
Vậy ......
6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x=17.\)
Vậy ...........
\(\)
Ta có:\(\dfrac{x-1}{79}=\dfrac{-28}{1-x}\)
\(\Rightarrow\left(x-1\right)\left(1-x\right)=-28.79\)
\(\Rightarrow\left(x-1\right).1-\left(x-1\right)x=-2212\)
\(\Rightarrow x-1-x^2+x=-2212\)
\(\Rightarrow x+x-x^2=-2211\)
\(\Rightarrow2x-x^2=-2211\)
\(\Rightarrow x\left(2-x\right)=-2211\)
...
\(\dfrac{x-1}{79}=\dfrac{-28}{1-x}\)
\(\Leftrightarrow\left(x-1\right)\left(1-x\right)=\left(-28\right).79\)
\(\Leftrightarrow x\left(1-x\right)-1\left(1-x\right)=-2212\)
\(\Leftrightarrow x-x^2-1-x=-2212\)
\(\Leftrightarrow-x^2-1=-2212\)
\(\Leftrightarrow-x^2=-2211\)
\(\Leftrightarrow x^2=2211\)
\(\Leftrightarrow x=\sqrt{2211}\)