Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
à sửa lại đề chút
\(a,\forall x\in R,x>3\Leftrightarrow x^2>9\)
a) 11+(15-x)=1
⇔15-x=-10
hay x=25
Vậy: x=25
b) 2x-35=15
⇔2x=50
hay x=25
Vậy: x=25
c) 2|x+5|=12
⇔|x+5|=6
\(\Leftrightarrow\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
Vậy: x∈{1;-11}
A/ \( 11+(15-x)=1\)\(\Leftrightarrow11+15-x=1\Leftrightarrow x=11+15-1=25\)
KL: ...........
B/ \(2x-35=15\Leftrightarrow2x=15+35=50\Leftrightarrow x=25\)
KL: ............
C/ \(2\left|x+5\right|=12\Leftrightarrow\left|x+5\right|=6\Leftrightarrow\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
KL: .............
\(a)13-\left(40-X\right)=35\)
\(\Leftrightarrow40-X=13-35\)
\(\Leftrightarrow40-X=-22\)
\(\Leftrightarrow X=40-\left(-22\right)\)
\(\Leftrightarrow X=62\)
Vậy \(X=62\)
\(b)14-3x\left(5-X\right)=8\)
\(\Leftrightarrow3x\left(5-X\right)=14-8\)
\(\Leftrightarrow3x\left(5-X\right)=6\)
\(\Leftrightarrow5-X=6:3\)
\(\Leftrightarrow5-X=3\)
\(\Leftrightarrow X=5-3\)
\(\Leftrightarrow X=2\)
Vậy \(X=2\)
\(c)\left(3X-2\right)x\left(5+X\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)
Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)
\(d)\left(3X-6\right)x3=3^4\)
\(\Leftrightarrow\left(3X-6\right)x3=81\)
\(\Leftrightarrow3X-6=81:3\)
\(\Leftrightarrow3X-6=27\)
\(\Leftrightarrow3X=27+6\)
\(\Leftrightarrow3X=33\)
\(\Leftrightarrow X=33:3\)
\(\Leftrightarrow X=11\)
Vậy\(X=11\)
\(\dfrac{2}{x}=\dfrac{x}{8}\Rightarrow x^2=16=4^2=\left\{{}\begin{matrix}+4\\-4\end{matrix}\right.\)\(A=\dfrac{4116-14}{10290-35}=\dfrac{4102}{10255}=\dfrac{2}{3}\)