3^x-2.5=135
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C) x-135=315-450
X-135=-135
X=0
D) 346+(210-x)=556
X-210=346-556
X-210=-210
X=0
Lời giải:
a. $145-135(x-2)^2=10$
$135(x-2)^2=145-10=135$
$(x-2)^2=135:135=1=1^2=(-1)^2$
$\Rightarrow x-2=1$ hoặc $x-2=-1$
$\Rightarrow x=3$ hoặc $x=1$
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b.
$x\in Ư(36); x\geq 12$
$\Rightarrow x\in\left\{12; 18; 36\right\}$
c.
$x-1=B(9)\Rightarrow x-1\in\left\{0; 9; 18; 27; 36; 45;54;.....\right\}$
$\Rightarrow x\in\left\{1; 10; 19; 28; 37;46; 55;....\right\}$
Mà $25< x< 50$ nên $x\in\left\{28; 37; 46\right\}$
TL :
\(3\frac{2}{5}\)x 2 x X = \(\frac{3}{5}\)
\(\frac{17}{5}\)x 2 x X = \(\frac{3}{5}\)
\(\frac{34}{5}\)x X = \(\frac{3}{5}\)
X = \(\frac{34}{5}\): \(\frac{3}{5}\)
X = \(\frac{34}{3}\)
\(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{5}=0\\x^2+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{1}{5}\\x^2=-\dfrac{1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{2}{15}\)
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)
\(\frac{4}{7}+x\times2=\frac{5}{3}\)
\(x\times2=\frac{5}{3}-\frac{4}{7}\)
\(x\times2=\frac{23}{21}\)
\(x=\frac{23}{21}:2\)
\(x=\frac{23}{42}\)
\(\frac{4}{7}+x\times2=\frac{5}{3}\)
\(x\times2=\frac{5}{3}-\frac{4}{7}\)
\(x\times2=\frac{23}{21}\)
\(x=\frac{23}{21}:2\)
\(x=\frac{23}{42}\)
Vậy x = 23/42
Ta có: \(3^{x-2}\cdot5=135\)
=>\(3^{x-2}=\frac{135}{5}=27=3^3\)
=>x-2=3
=>x=3+2=5
sai đề, có phải không