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1 a x=4
b x=-4
c x=-7
d x=3
e x=10
g x=60
h x=36
i x=16
2a 1,2,3,4,5,6,7,8,9
b 1,2,3,4,5,6,7,8,9.........
c rỗng
3a 0
b 0
c10
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
\(2x+4⋮x-1\Rightarrow2\left(x-1\right)+6⋮x-1\)
\(\Rightarrow6⋮x-1\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
Vậy...........................................
\(2x^2+\left(-3\right)^2=41\)
\(\Rightarrow2x^2=41-9=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
\(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Rightarrow2x-10-3x-21=14\)
\(\Rightarrow2x-3x=14+21+10\)
\(\Rightarrow-x=45\Rightarrow x=-45\)
\(-7\left(5-x\right)-2\left(x-10\right)=15\)
\(\Rightarrow-35+x-2x+20=15\)
\(\Rightarrow x-2x=15-20+35\)
\(\Rightarrow-x=30\Rightarrow x=-30\)
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
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)