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<=> [x2-(36-1-35)3-15]3=1
<=> (x2-15)3=1
<=> x2-15=1
<=> x2=16
<=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
a/ 2x . 7 = 224
2x = 224 : 7
2x = 32
2x = 25
x = 5.
b/ 32x + 1 . 11 = 2673
32x + 1 = 2673 : 11
32x + 1 = 243
32x + 1 = 35
32x = 35 - 1
32x = 34
2x = 4
x = 4 : 2
x = 2.
c/ (3x + 5)2 = 289
(3x + 5)2 = 172
3x + 5 = 17
3x = 17 - 5
3x = 12
x = 12 : 3
x = 4.
d/ x . (x2)3 = x5
x1 . x5 = x5
x6 = x5
=> x = 0 hoặc x = 1
a) 2x x 7=224
2x=224:7
2x=32
2x=25
=> x=5
Vậy x=5
b) 32x+1 x 11=2673
32x+1=2673:11
32x+1=243
32x+1=35
=> 2x+1=5
x=(5-1):2
x=2
Vậy x=2
c) (3*x+5)2=289
(3*x+5)2=172
=> 3*x+5=17
x=(17-5):3
x=4
Vậy x=4
d) x.(x2)3=x5
x.x6=x5
x=x5:x6
x=x-1
{x2-[36(64-63)3-35]-15}3=1
{x2-[36.13-35]-15}3=1
{x2-[36.1-35]-15}3=1
{x2-[36-35]-15}3=1
{x2-1-15}3=1
{x2-1-15}3=13
=>x2-1-15=1
x2=1+1+15
x2=17
x2=căn của 7
=>x=căn 7
Vậy x=căn 7
tk nha
<=>{x2-[62(64-63)3-35]3-15}3=1
<=>[x2-(62.13-35)3-15]3=1
<=>[x2-(36-35)3-15]3=1
<=>(x2-1-15)3=1
<=>(x2-16)=\(\sqrt{1}=1\)
=>x2=17=>x=\(\sqrt{17}\)
\(2^{x+1}.3^y=12^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)
\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)
\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-\left(8^2-\left(8+1\right).\left(8-1\right)\right)^3-\left(6+1\right).\left(6-1\right)\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-\left(8^2-8^2+1\right)^3-6^2+1\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-\left[6^2-1-6^2+1\right]^3-5.3\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-5.3\right\}^3=1\)
\(\Leftrightarrow x^2-15=1\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)-7.5\right]^3-5.3\right\}^3=1\)
\(\Rightarrow x^2-\left[6^2-\left(64-63\right)^3-35\right]^3-5.3=1\)
\(\Rightarrow x^2-\left(36-1^3-35\right)^3-15=1\)
\(\Rightarrow x^2-0^3-15=1\)
\(\Rightarrow x^2=1+15\)
\(\Rightarrow x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
=> (n-2)^4 - (n-2)^2 = 0
=> (n-2)^2.[(n-2)^2-1] = 0
=> (n-2)^2.(n-3).(n-1) = 0
=> n-2=0 hoặc n-3=0 hoặc n-1=0
=> n=1 hoặc n=2 hoặc n=3
Vậy ............
Tk mk nha
\(\left(n-2\right)^2=\left(n-2\right)^4\)
\(\Rightarrow\orbr{\begin{cases}n-2=0\\n-2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}n=2\\n=3\end{cases}}\)
vậy_____
\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3-5.3\right\}^3=1\)
\(\Rightarrow\left\{x^2-\left(36-1^3-35\right)^3-15\right\}^3=1\)
\(\Rightarrow x^2-\left(0^3-15\right)^3=1\)
\(\Rightarrow x^2-\left(-3375\right)=1\)
\(\Rightarrow x^2=-3374\)
\(\Rightarrow x\in\varnothing\)
{ x2 - [ 62 - ( 82 - 9.7)3 - 7.5]3 - 5.3 }3 = 1
{ x2 + [ 36 - (64 - 63)3 - 35]3 - 15}3 = 1
[ x2 - ( 36 - 13 - 35 ) - 15 ]3 = 1
[ x2 - ( 36 - 1 - 35 ) - 15]3 = 1
[ x2 - ( 35 - 35 ) - 15]3 = 1
[ x2 - 0 - 15]3 = 1
( x2 - 15 )3 = 1
<=> ( x2 - 15)3 = 13
=> x2 - 15 = 1
<=> x2 = 16
=> x = 4