\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}\)

\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)

\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)

\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)

\(\Leftrightarrow x=\dfrac{42}{22}\)

\(\Leftrightarrow x=\dfrac{21}{11}\)

\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)

\(\Leftrightarrow x=\dfrac{29}{28}\)

\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)

\(x=\dfrac{5}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow x=\dfrac{5}{4}\times8\)

\(\Leftrightarrow x=\dfrac{40}{4}\)

\(\Leftrightarrow x=10\)

1) 

=a^4+2a^2+1-a^2

=(a^2+1)^2-a^2

=(a^2-a+1)(a^2+a+1)

2)

=a^4+4b^4-4a^2b^2

=(a^2+2b^2)^2-4a^2b^2

=(a^2-2ab+2b^2)(a^2+2ab+2b^2)

3)

=(8x^2+1)^2-16x^2

=(8x^2-4x+1)(8x^2+4x+1).

4)

=x^5+x^4+x^3-x^3+1

=x^2(x^2+x+1)-(x-1)(x^2+x+1)

=(x^2-x+1)(x^2+x+1)

5).

=x^7-x+x^2+x+1

=x(x^6-1)+x^2+x+1

=x(x^3-1)(x^3+1)+x^2+x+1

=x(x-1)(x^2+x+1)(x^3+1)+x^2+x+1

=(x^2+x+1)[(x^2-x)(x^3+1)+1]

6)

=x^8-x^2+x^2+x+1

=x^2(x-1)(x^2+x+1)(x^3+1)+x^2+x+1

Xong nhóm x^2+x+1 vào.

7)

=x^4-(2x-1)^2

=(x^2-2x+1)(x^2+2x-1)

8)

=(a^8+b^8)^2-a^8b^8

=(a^8-a^4b^4+b^8)(a^8+a^4b^4+b^8).

`x/8 = 3/4 +(-5/8)`

`=>x/8 = 6/8 +(-5/8)`

`=>x/8 = 1/8`

`=>x=1`

`-----`

`x/12 =3/4 +(-2/3)`

`=>x/12 = 9/12 + (-8/12)`

`=> x/12=1/12`

`=>x=1`

`----`

`1+11/13=24/x`

`=> 13/13 +11/13=24/x`

`=> 24/13 =24/x`

`=>x=13`

`----`

`x/6 -3/4=1/12`

`=>x/6 = 1/12 +3/4`

`=>x/6 = 1/12 + 9/12`

`=>x/6 = 10/12`

`=>x/6= 5/6`

`=>x=5`

\(a)\)

\(\dfrac{x}{8}=\dfrac{3}{4}+\dfrac{-5}{8}\)

\(\Rightarrow\dfrac{x}{8}=\dfrac{1}{8}\)

\(\Rightarrow x.8=1.8\)

\(\Rightarrow x.8=8\)

\(\Rightarrow x=8:8\)

\(\Rightarrow x=1\)

\(b)\)

\(\dfrac{x}{12}=\dfrac{3}{4}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}\)

\(\Rightarrow x.12=1.12\)

\(\Rightarrow x.12=12\)

\(\Rightarrow x=12:12\)

\(\Rightarrow x=1\)

\(c)\)

\(1+\dfrac{11}{13}=\dfrac{24}{x}\)

\(\Rightarrow\dfrac{24}{13}=\dfrac{24}{x}\)

\(\Rightarrow x.24=24.13\)

\(\Rightarrow x.24=312\)

\(\Rightarrow x=312:24\)

\(\Rightarrow x=13\)

\(d)\)

\(\dfrac{x}{6}-\dfrac{3}{4}=\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{1}{12}+\dfrac{3}{4}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{5}{6}\)

\(\Rightarrow x.6=5.6\)

\(\Rightarrow x.6=30\)

\(\Rightarrow x=30:6\)

\(\Rightarrow x=5\)

a: \(x+6\dfrac{1}{8}=8\)

=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)

=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)

b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)

=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)

c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)

=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)

=>\(x\cdot1=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}\)

1, a⁴ + a² + 1 

= a⁴ + 2a² + 1 - a² 

= (a²)² + 2a² + 1 - a² 

= (a² + 1)² - a² 

= (a² + 1 - a)(a² + 1 + a)

2, a⁴ + 4b⁴ 

= (a²)² + 2. a² . b² + (2b)² - a² . b² 

= (a² + 2b)² - (ab)² 

= (a² + 2b - ab)(a² + 2b + ab)

3, 64x⁴ + 1 

= (8x²)²​ + 16x²​ + 1 - 16x²​ 

= (8x² + 1)²​ - (4x)²​ 

= (8x² + 1 - 4x)(8x² + 1 + 4x)

4, x⁵ + x⁴ + 1 

= x⁵ + x⁴ + x³ - x³ - x² - x + x + x² + 1 

= (x⁵ + x⁴ + x³) - (x³ + x² + x) + (x + x² + 1)

= x³(x² + x + 1) - x(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x³ - x + 1)

5, x⁷ + x² + 1 

= x⁷ – x + x² + x + 1

= x(x⁶ – 1) + (x² + x + 1) 

= x(x³ – 1)(x³ + 1) + (x² + x + 1)

= x(x³ + 1)(x – 1) (x² + x + 1) + (x² + x + 1) 

= (x² + x + 1)[ x(x³ + 1)(x – 1) + 1]

= (x² + x + 1)(x⁵ – x⁴ + x³ – x² + x – 1)

6, x⁸ + x + 1 

= x⁸ + x⁷ + x⁶ - x⁷ - x⁶ - x⁵ + x⁵ + x⁴ + x³ - x⁴ - x³ - x² + x² + x + 1

= (x⁸ + x⁷ + x⁶) -  (x⁷ + x⁶ + x⁵) + (x⁵ + x⁴ + x³ ) - (x⁴ + x³ + x²) + (x² + x + 1)

= x⁶(x² + x + 1) - x⁵(x² + x + 1) + x³(x² + x + 1) - x²(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁶ - x⁵ + x³ - x² + 1)

7, x⁴ - 4x² + 4x - 1 

= x⁴ - (4x² - 4x + 1)

= (x²)² - (2x - 1)²

= (x² - 2x + 1)(x² + 2x - 1)

= (x - 1)² (x² + 2x - 1)

8, a¹⁶ + a⁸b⁸ + b¹⁶

=  (a¹⁶ + 2a⁸b⁸ + b¹⁶) - a⁸b⁸ 

= (a⁸ + b⁸)² - (a⁴b⁴)²

= (a⁸ + b⁸ - a⁴b⁴)(a⁸ + b⁸ + a⁴b⁴)

= (a⁸ + b⁸ - a⁴b⁴)[(a⁸ + b⁸ + 2a⁴b⁴) - a⁴b⁴]

= (a⁸ + b⁸ - a⁴b⁴)[(a⁴ + b⁴)² - (a²b²)²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)(a⁴ + b⁴ + a²b²)

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)[(a⁴ + b⁴ + 2a²b²) - a²b²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)[(a² + b²) - (ab)²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)(a² + b² - ab)(a² + b² + ab)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)

`4.(3-x)^10=9(3-x)^8`

`=>(3-x)^{8}(4(3-x)^2-9)=0`

`=>` $\left[ \begin{array}{l}3-x=0\\(3-x)^2=\dfrac{9}{4}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\3-x=\dfrac{3}{2}\\3-x=-\dfrac{3}{2}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\\x=\dfrac{9}{2}\end{array} \right.$

Vậy `x=3` hoặc `x=3/2` hoặc `x=9/2`

`b,(x-1)^{x+3}=(x-1)^{x+1}`

`=>(x-1)^{x+1}[(x-1)^2-1]=0`

`=>` $\left[ \begin{array}{l}x-=1\\x-=-1\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=2\\x=1\\x=0\end{array} \right.$

Vậy `x=0` hoặc `x=1` hoặc `x=2`

a) 4(3 - x)¹⁰ = 9(3 - x)⁸

=> (x - 3)⁸.[4(x - 3)² - 9) = 0

<=> (x - 3)⁸.(2x - 6 - 3)(2x - 6 + 3) = 0

<=> (x - 3)⁸(2x - 9)(2x - 3) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\2x-9=0\\2x-3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{9}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy S = {3;9/2; 3/2}

b) (x - 1)^{x + 3} = (x - 1)^{x  + 1}

<=> (x - 1)^{x + 1}.[(x - 3)² - 1) = 0

<=>(x - 1)^{x + 1}.(x - 3 - 1)(x - 3 + 1) = 0

<=> (x - 1)^{x + 1}.(x - 4)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=4\\x=2\end{matrix}\right.\)

b) Ta có: \(\left(x-1\right)^{x+3}=\left(x-1\right)^{x+1}\)

\(\Leftrightarrow\left(x-1\right)^{x+3}-\left(x-1\right)^{x+1}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+1}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+1}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy: S={0;1;2}

1:

a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)

b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)

nên 2 p/s ko bằng nhau