Tìm x biết:
a) (2x - 1)3 = - 27
b) (x - 3)10 = (x - 3)30
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Lời giải:
a. $121-3(x-5)=6$
$3(x-5)=121-6=115$
$x-5=115:3=\frac{115}{3}$
$x=\frac{115}{3}+5=\frac{130}{3}$
b.
$2x-138=2^3.3^2=72$
$2x=72+138=210$
$x=210:2=105$
c.
$x-3\vdots 7$
$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$
Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$
d.
$27\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$
$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$
a ) 121-3.(x - 5 ) = 6
3.(x-5) = 121 -6
3. (x-5)=115
x-5 = 115:3
x-5=35
x=35+5
x = 40
b) 2x - 138 = 2'3. 3'2
2x -138=8.9
2x-138=72
2x=72+138
2x=210
x=210:2
x=105
c) theo bài ra : x-3 ∈ B(7)
ta có B(7)=(0,7,14,21,28,35,49,56,...)
=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)
=) x ∈( 3 , 10,17,24,31,38,42,58,..)
mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )
vậy x ∈(17,24,31,38,42)
Lời giải:
a.
$(25-2x)^3:5-3^2=4^2$
$(25-2x)^3:5=4^2+3^2=25$
$(25-2x)^3=25.5=5^3$
$\Rightarrow 25-2x=5$
$\Rightarrow 2x=20$
$\Rightarrow x=10$
b.
$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$
$\Rightarrow 3^x=3^{14}$
$\Rightarrow x=14$
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a) \(\left(2x-1\right)^3=-27\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow2x-1=-3\)
\(\Leftrightarrow2x=-3+1\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\)
b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)
\(\Leftrightarrow\left(x-3\right)^{10}=\left[\left(x-3\right)^{10}\right]^3\)
\(\Leftrightarrow x-3=\left(x-3\right)^3\)
giải ra x = 4 ; x = 2; x = 3
\(\text{a) }\left(2x-1\right)^3=-27\\ \Leftrightarrow\left(2x-1\right)^3=-3^3\\ \Leftrightarrow2x-1=-3\\ \Leftrightarrow2x=-2\\ \Leftrightarrow x=-1\\ \text{Vậy }x=-1\\ \)
\(\text{b) }\left(x-3\right)^{10}=\left(x-3\right)^{30}\\ \Leftrightarrow\left(x-3\right)^{10}-\left(x-3\right)^{30}=0\\ \Leftrightarrow\left(x-3\right)^{10}\left[1-\left(x-3\right)^{20}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^{10}=0\\1-\left(x-3\right)^{20}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{20}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ \text{Vậy }x=3\text{ hoặc }x=4\)