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a) 7/2 - x = 0,75
x = 7/2 - 0,75
x = 11/4
b) 2 |x - 1/3 | + 1/ 5 = 1
2 | x - 1/3 | = 1 - 1,5
2| x - 1/3 | = - 0,5
x - 1/3 = -0,5 : 2 => x = -0,25 = -1/4
x - 1/3 = 0,5 : 2 => x = 0,25 = 1/4
Mình nghĩ câu c bạn nên suy nghĩ nhé , chúc bạn may mắn
\(\frac{x+1}{125}+\frac{x+2}{124}+\frac{x+3}{123}+\frac{x+4}{122}+\frac{x+146}{5}=0\)
\(\left(\frac{x+1}{125}+1\right)+\left(\frac{x+2}{124}+1\right)+\left(\frac{x+3}{123}+1\right)+\left(\frac{x+4}{122}+1\right)+\left(\frac{x+146}{5}-4\right)=0\)
\(\frac{x+126}{125}+\frac{x+126}{124}+\frac{x+126}{123}+\frac{x+126}{122}+\frac{x+126}{5}=0\)
\(\left(x+126\right).\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)=0\)
vì \(\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)\ne0\)nên x + 126 = 0 \(\Rightarrow\)x = -126
Bài giải
a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)
\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)
\(\frac{2}{7}x=-\frac{5}{4}\)
\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)
\(x=-\frac{35}{8}\)
b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)
\(6x=-\frac{8}{125}-\frac{2}{5}\)
\(6x=-\frac{58}{125}\)
\(x=-\frac{58}{125}\text{ : }6\)
\(x=\frac{-29}{375}\)
c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)
Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
\(5^{x+2}=125\Rightarrow5^x.5^2=125\Rightarrow5^x=125:5^2\Rightarrow5^x=5\Rightarrow5^x=5^1\Rightarrow x=1\)
\(\left|x+3\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x+3\right|=\frac{1}{3}\)
\(\Rightarrow\begin{cases}x+3=\frac{1}{3}\\x+3=-\frac{1}{3}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\)
a) 5x+2 = 125
=> 5x+2 = 52
=> x + 2 = 2
=> x = 2 - 2
=> x = 0
b) \(\left|x+3\right|-\frac{2}{3}=1\)
=> \(\left|x+3\right|=1+\frac{2}{3}\)
=> \(\left|x+3\right|=\frac{5}{3}\)
=> \(\left[\begin{array}{nghiempt}x+3=\frac{5}{3}\\x+3=-\frac{5}{3}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{4}{3}\\x=-\frac{14}{3}\end{array}\right.\)