1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

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: x^3=7^3

=>x^3=343

=>\(x=\sqrt[3]{343}=7\)

b: x^3=27

=>x^3=3^3

=>x=3

c: x^3=125

=>x^3=5^3

=>x=5

d: (x+1)^3=125

=>x+1=5

=>x=4

e: (x-2)^3=2^3

=>x-2=2

=>x=4

f: (x-2)^3=8

=>x-2=2

=>x=4

h: (x+2)^2=64

=>x+2=8 hoặc x+2=-8

=>x=6 hoặc x=-10

j: =>x-3=2 hoặc x-3=-2

=>x=1 hoặc x=5

k:

9x^2=36

=>x^2=36/9

=>x^2=4

=>x=2 hoặc x=-2

l:

(x-1)^4=16

=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

1.

a) \(2^x=128\)

\(2^x=2^7\)

\(=>x=7\)

b) \(8^{x-1}=64\)

\(8^{x-1}=8^2\)

\(=>x-1=2\)

\(x=2+1\)

\(=>x=3\)

c) \(3+3^x=30\)

\(3^x=30-3\)

\(3^x=27=3^3\)

\(=>x=3\)

d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?

e) \(3^2.x=3^5\)

\(x=3^5:3^2\)

\(=>x=3^3=27\)

f) \(\left(2x-1\right)^3=343\)

\(\left(2x-1\right)^3=7^3\)

\(=>2x-1=7\)

\(2x=7+1\)

\(2x=8\)

\(x=8:2\)

\(=>x=4\)

\(#Wendy.Dang\) 

a,\(2^x\)=128           b,\(8^{x-1}\)=64              c,3+\(3^x\)=30           d,x+2=64

\(2^7\)=128               \(8^{x-1}\)=\(8^2\)                 \(3^x\)=30-3                  x=64-2

=>x=7              =>x-1=2                  \(3^x\)=27                      x=62

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

                                                     =>x=3

e,\(3^2\).x=\(3^5\)                             f,(2x-\(1^3\))=343

x=\(3^5\):\(3^2\)                                 2x=1+343

x=27                                     2x=344

                                               x=344:2

                                               x=172

\(x^3+1\)

\(=x^3+1^3\)

\(=\left(x+1\right)\left(x^2-x+1\right)\)

______

\(8+x^3\)

\(=2^3+x^3\)

\(=\left(2+x\right)\left(4-2x+x^2\right)\)

______

\(27x^3-64y^3\)

\(=\left(3x\right)^3-\left(4y\right)^3\)

\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)

______

\(\dfrac{x^3}{64}-\dfrac{1}{125}\)

\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)

\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)

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

x^3+8=(x+2)(x^2-2x+4)

27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)

\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)

x³+1= (x+1)(x²-x+1)

8+x³=2³+x³=(2+x)(4-2x+x²)

27x³-64y³= (3x)³-(4y)³= (3x-4y)(9x²+12xy+16y²)

\(\dfrac{x^3}{64}\)-\(\dfrac{1}{125}\)(sửa đề)=(x/4)³-(1/5)³

=(x/4-1/5)(x/16+x/20+1/25)

\(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow x-3=\left(x-3\right)^2\)

\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

___________

\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

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

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

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

C nha

Chúc bn học tốt!

Sửa lại nha, mk nhìn nhầm đề, -x = 1 thì x = -1 vậy là D nha

Chúc bn học tốt!

D.-1

Nhớ tick cho mình nhé!

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

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

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

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)