3 x 2 ( x - 10 ) = 64
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1: -5>-10
=>x-5>x-10
2: -4>-8
=>x-4>x-8
3: 2>-6
=>x+2>x-6
4: -3<7
=>x-3<x+7
5: 5<8
=>x+5<x+8
6: 10>7
=>x+10>x+7
1/
$(x-1)^{x+10}=(x-1)^{x+8}$
$\Rightarrow (x-1)^{x+10}-(x-1)^{x+8}=0$
$\Rightarrow (x-1)^{x+8}(x^2-1)=0$
$\Rightarrow (x-1)^{x+8}=0$ hoặc $x^2-1=0$
Nếu $(x-1)^{x+8}=0\Rightarrow x-1=0\Rightarrow x=1$
Nếu $x^2-1=0\Rightarrow x^2=1=1^2=(-1)^2\Rightarrow x=1$ hoặc $x=-1$
Vậy $x=1$ hoặc $x=-1$
2/
$1^3+2^3+3^3+...+10^3=(x+1)^2$
Ta có công thức quen thuộc:
$1^3+2^3+...+n^3=(1+2+...+n)^2=\frac{[n(n+1)]^2}{4}$
Bạn có thể xem cm tại đây:
https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/
Khi đó:
$1^3+2^3+...+10^3=(x+1)^2$
$\Rightarrow \frac{[10(10+1)]^2}{4}=(x+1)^2$
$\Rightarrow 3025=(x+1)^2$
$\Rightarrow x+1=55$ hoặc $x+1=-55$
$\Rightarrow x=54$ hoặc $x=-56$
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
a, 410. 230=220.230=250
b,925.274.813= 350.312.312=374
Tương tự các câu khác....
a, 410.230 = (22)10.230 = 220.230 = 250
b, 925.274.813 = (32)25.(33)4.(34)3 = 350.312.312 = 374
c, 2550.1255 = (52)50.(53)5 = 5100.515 = 5115
d, 643.48.164 = (26)3.(22)8.(24)4 = 218.216.216 = 250
e, 38 : 36 = 32
210 : 83 = 210 : (23)3 = 210 : 29 = 2
127 : 67 = (12 : 6)7 = 27
@Dương Tuyết Mai
2x = 64
⇒ 2x = 26
⇒ x = 6.
7x = 710 : 75
⇒ 7x = 75
⇒ x = 5.
3x = 315 : 243
⇒ 3x = 315 : 35
⇒ 3x = 310
⇒ x = 10.
\(2^x=64=2^6\Rightarrow x=6\\ ---\\ 7^x=7^{10}:7^5=7^5\Rightarrow x=5\\ ----\\ 3^x=3^{15}:243=3^{15}:3^5=3^{10}\Rightarrow x=10\)
a) \(\left(\frac{2x}{3}-3\right)\div\left(-10\right)=\frac{2}{5}\)
\(\Rightarrow\left(\frac{2x}{3}-\frac{9}{3}\right).-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow\frac{2x-9}{3}.-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow\frac{9-2x}{30}=\frac{2}{5}\Rightarrow\left(9-2x\right).5=60\)
\(\Rightarrow45-10x=60\Rightarrow-10x=15\Rightarrow x=-\frac{15}{10}=-1,5\)
b)\(\left(x+5\right)^3=64\)
\(\Rightarrow\left(x+5\right)^3=4^3\)
\(\Rightarrow x+5=4\Rightarrow x=-1\)
A, 2x/3-3=-4
2x/3=-1
2x=-3
X=-3/2
B,(x+5)^3=4^3
=> x+5=4
X=-1
k nha
\(a.3^x.2=54\Leftrightarrow3^x=27\Leftrightarrow3^x=3^3\Leftrightarrow x=3\)
\(c.\left(2x+1\right)^3=64\Leftrightarrow\left(2x+1\right)^3=4^3\)
\(\Leftrightarrow2x+1=4\Leftrightarrow x=\dfrac{3}{2}\)
a: =>3^x=27
=>x=3
b: =>x(x^9-1)=0
=>x=0 hoặc x=1
c: =>2x+1=4
=>2x=3
=>x=3/2
Ta có: \(2^x=\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)
\(\Leftrightarrow2^x=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot31}{2\cdot\left(2\cdot3\cdot4\cdot...\cdot31\right)\cdot64}\)
\(\Leftrightarrow2^x=\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{128}\)
\(\Leftrightarrow2^x=\dfrac{1}{2^6}\)
\(\Leftrightarrow2^{x+6}=1\)
\(\Leftrightarrow x+6=0\)
hay x=-6
Vậy: x=-6
`1/4 . 2/6 . 3/8 ... . 30/62 .31/64 =2^x`
`-> (1.2.3....30.31)/(4.6.8....62.64)=2^x`
`-> (1.(2.3...31))/(2.(2.3.4...31).32)=2^x`
`-> 1/(2.32)=2^x`
`-> 1/64=2^x`
`-> 1/(2^6)=2^x`
`-> x=-6`.
9x2 - 4 - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0
<=> ( 3x - 2 )( 2x - 3 ) = 0
<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)
x3 + 64 + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 + 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 2x + 13 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x2 - 2x + 13 = ( x2 - 2x + 1 ) + 12 = ( x - 1 )2 + 12 ≥ 12 > 0 ∀ x )
( x - 3 )( x2 + 4x + 9 ) + 2( x2 - 9 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 + 2x + 6 - 10 ) = 0
<=> ( x - 3 )( x2 + 6x + 5 ) = 0
<=> ( x - 3 )( x + 1 )( x + 5 ) = 0
<=> x = 3 hoặc x = -1 hoặc x = -5
<=> ( x - 3 )(
\(6\left(x-10\right)=64\Leftrightarrow x-10=\dfrac{32}{3}\Leftrightarrow x=\dfrac{62}{3}\)
3 . 2 .(x-10) = 64
=> 6 . (x-10) = 64
=> x - 10 = \(\dfrac{64}{6}=\dfrac{32}{3}\)
\(\Rightarrow x=\dfrac{32}{3}+10=\dfrac{62}{3}\)