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\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)
a: (x-1)(x+2)(-x-3)=0
=>(x-1)(x+2)(x+3)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
b: (x-7)(x+3)<0
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
Ý là đề vầy chứ gì:
\(5^x.5^{x+1}.5^{x+2}=10^{18}:2^{18}\)
⇔\(5^{3x+3}=5^{18}\)
⇔\(3x+3=18\)
⇔\(x=5\)
Vậy x=5
\(78+23\cdot81-69=78+1863-69\)
\(=1941-69\)
\(=1872\)
\(78+23\cdot\left(81-69\right)=78+23\cdot12\)
\(=78+276\)
\(=354\)
\(\left(78+23\right)\cdot81-69=101\cdot81-69\)
\(=8181-69\)
\(=8112\)
\(\left(78+23\right)\cdot\left(81-69\right)=101\cdot12\)
\(=1212\)
78 x 31 + 78 x 24 + 55 x 22
= 78 x ( 31 + 24 ) + 55 x 22
= 78 x 55 + 55 x 22
= 55 x ( 78 + 22 )
= 55x 100
= 5500
\(=78\cdot64-78\cdot42+78\cdot78\)
\(=78\left(64-42+78\right)\)
\(=78\cdot100\)
\(=7800\)
(x - 78) x 26 = 0
x - 78 = 0 : 26
x = 0 + 78
x = 78
(X - 78) * 26 =0
(X - 78) = 0 *26
(X - 78) =0
X = 78 + 0
X =78