(13x-12²):5=5

13x-12² = 5 . 5 

13x-12² = 25

       13x = 25 + 12²

       13x = 169

            x = 169 : 13

            x = 13

Vậy x = 13 

cảm ơn bn nhìuuu

(13x-12^2):5=5

=)) (13 x- 12^2) =5×5

=)) (13×-12^2)= 25

=))13 x = 25+144

=)) x = 169:13

=))x= 13

Thay x=2 và y=-1 vào (d),ta được:

2(m^2-5m+1)+2m-6=-1

=>2m^2-10m+2+2m-6+1=0

=>2m^2-8m-3=0

=>\(m=\dfrac{4\pm\sqrt{22}}{3}\)

=41/10

x - 1/2 = 18/5

x = 18/5 + 1/2

x = 41/10

\(x-\dfrac{1}{2}=\dfrac{3}{5}\times6\)

\(x-\dfrac{1}{2}=\dfrac{18}{5}\)

\(x=\dfrac{18}{5}+\dfrac{1}{2}\)

\(x=\dfrac{41}{10}\)

Vì 1⁵= 1⁶ nên x - 5 = 1

x= 1+5

x=6

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\left(x-5\right)^5=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^5.\left(x-6\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^5=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{KOH}=0,3.x\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,06\left(mol\right)\end{matrix}\right.\)

\(n_{BaCO_3}=\dfrac{9,85}{197}=0,05\left(mol\right)\)

PTHH: Ba(OH)₂ + CO₂ --> BaCO₃ + H₂O

_______0,06--->0,06----->0,06

2KOH + CO₂ --> K₂CO₃ + H₂O

0,3x->0,15x---->0,15x

K₂CO₃ + CO₂ + H₂O --> 2KHCO₃

0,15x->0,15x

BaCO₃ + CO₂ + H₂O --> Ba(HCO₃)₂

0,01--->0,01

=> 0,06 + 0,3x + 0,01 = 0,1

=> x = 0,1

B=2+2²+2³+...+2¹⁰⁰

2B=2²+2³+2⁴+...+2¹⁰¹

2B-B=(2²+2³+2⁴+...+2¹⁰¹)-(2+2²+2³+...+2¹⁰⁰)

B=2¹⁰¹-2

Theo như đề bài thì B+2=2^X mà B=2¹⁰¹-2

Vậy B+2=2¹⁰¹-2+2=2¹⁰¹=2^x

Suy ra x=101 

Đáp số 101

\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)

<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)

<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)

<=> \(-\frac{1}{3}x=\frac{29}{12}\)

<=> \(x=-\frac{29}{4}\)

\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x=-\frac{5}{6}\)

<=> \(x=5\)

học tốt

Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$

$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
 

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)