#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)

h)Tương tự các câu trên

i) x = 0

k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)

l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)

=> x + 1 = 3 => x = 2

x.(x+1)=0

suy ra x=0 hoac x+1=0

                               x=0-1

                              x=-1

vay x=0 hoac  x=-1

mấy câu sau cũng làm tương tự

\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=0\)

dễ thế mà ko biết làm ak!

1, x² = 0

=> x=0

2,x²=1

=> x= 1 hoặc x=-1

3,x²=3

=>\(x=\sqrt{3}\)

4,x²=6

=>\(x=\sqrt{6}\)

5,x²=7

=>\(x=\sqrt{7}\)

a)

Điều kiện: $x\ne-2,-5,-10,-17.$

$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow x=15.$

b)

Điều kiện: $x\ne1,3,8,20.$

$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$

$\Leftrightarrow\dfrac1{x-1}=\dfrac34$

$\Leftrightarrow4=3(x-1)$

$\Leftrightarrow3x=7$

$\Leftrightarrow x=\dfrac73.$

\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\)  \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\)  \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\) \(0\)

a) 3.x-1⁴ =81

3.x = 81+1

3.x = 82

x = 82:3

x=82/3

Vậy x = 82/3

b) (x+1)⁵= -32

(x+1)⁵= (-2)⁵

^{x+1 = -2}

x = -2-1

x = -3

Vậy x = -3

Giải:

a) \(\dfrac{1}{4}+x-\dfrac{1}{4}x=\dfrac{3}{4}\)

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

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

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy ...

b) \(\left|x^2-2x\right|+\left|x\right|=0\)

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

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

\(\Leftrightarrow x=0\)

Vậy ...

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

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=x\\3x^2-2x=-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2=3x\\3x^2=x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-3x=0\\3x^2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x\left(x-1\right)=0\\x\left(3x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

Cảm ơn bn. Bn có thể giúp mk 2 p cuối ko???