Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\) phải k bạn nhỉ? `11/8` k có bậc lũy thừa nào `=5` á.
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{27}{8}\)
`=>`\(\left(\dfrac{x}{2}-1\right)^3=\left(-\dfrac{3}{2}\right)^3\)
`=>`\(\dfrac{x}{2}-1=-\dfrac{3}{2}\)
`=>`\(\dfrac{x}{2}=-\dfrac{3}{2}+1\)
`=>`\(\dfrac{x}{2}=-\dfrac{1}{2}\)
`=> x=1`
Vậy, `x=1`
`b)`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\0,75-1\dfrac{1}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\-\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\-3x\cdot100=2\cdot75\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x\cdot100=150\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x=1,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; -1/2}.`
Ta có: \(\dfrac{-4}{15}< \dfrac{5x-1}{18}< \dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{-48}{180}< \dfrac{10\left(5x-1\right)}{180}< \dfrac{75}{180}\)
Suy ra: \(-48< 10\left(5x-1\right)< 75\)
\(\Leftrightarrow10\left(5x-1\right)\in\left\{-40;-30;-20;-10;0;10;20;30;40;50;60;70\right\}\)
\(\Leftrightarrow5x-1\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7\right\}\)
\(\Leftrightarrow5x\in\left\{-3;-2;-1;0;1;2;3;4;5;6;7;8\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)(Vì x nguyên)
\(\Leftrightarrow x^2-2x+1-1=0\)
\(\Leftrightarrow\left(x-1\right)^2=1\Rightarrow\left(x-1\right)=\pm1\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(5^{x+1}+5^{x-1}=130\)
\(5^x\cdot5^1+5^x\div5^1=130\)
\(5^x\cdot5^1+5^x\cdot\dfrac{1}{5}=130\)
\(5^x\cdot\left(5+\dfrac{1}{5}\right)=130\)
\(5^x\cdot\dfrac{26}{5}=130\)
\(5^x=130\div\dfrac{26}{5}\)
\(5^x=130\cdot\dfrac{5}{26}\)
\(5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Mọi người còn câu trả lời nào khác không cứ trả lời đi mik tick cho
\(x.\left(x+1\right)=2+4+6+...+2500\)
=>\(x.\left(x+1\right)=\left[\left(2500-2\right):2+1\right].\left(2500+2\right):2\)
=>\(x.\left(x+1\right)=1250.2502:2\)
=>\(x.\left(x+1\right)=1250.1251\)
Vì: x và x+1 là 2 số tự nhiên liên tiếp x<x+1
1250 và 1251 là 2 số tự nhiên liên tiếp 1250<1251
=>x=1250
Áp dụng công thức tính dãy số ta có
(2005-2):2+1 . ( 2005 + 2):2 = X . (X+1)
1002,5 . 1003,5 = X .(X +1)
=> X = 1002,5
Áp dụng công thức tính dãy số :
(2005-2):2+1 . ( 2005 + 2):2 = X . (X+1)
1002,5 . 1003,5 = X .(X +1)
X = 1002,5
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
x(x+8)=20
=>\(x^2+8x-20=0\)
=>(x+10)(x-2)=0
=>\(\left[{}\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
\(x\)(\(x+8\)) = 20
\(x^2\) + 8\(x\) = 20
\(x^2\) + 8\(x\) - 20 = 0
(\(x^2\) + 10\(x\)) - (2\(x\) + 10) = 0
\(x\)(\(x+10\)) - 2(\(x+10\)) = 0
(\(x+10\))(\(x-2\)) = 0
\(\left[{}\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-10; 2}