\((2x-3)^3=(2x-3)^5\\\Rightarrow (2x-3)^3-(2x-3)^5=0\\\Rightarrow (2x-3)^3[1-(2x-3)^2]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-3\right)^3=0\\1-\left(2x-3\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x-3=1\\2x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x=4\\2x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

\(Vậy:x\in\left\{1;\dfrac{3}{2};2\right\}\)

NX: 2x+3; 5(2x+3) và 2(2x+3)  cùng dấu

+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)

=> 5(2x+3), 2(2x+3) \(\ge\)0

=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3

=> (2x+3)(5+2+1) = 16

=> 2x+3 = 2

=> 2x = -1

=> x = -1/2 (t/m)

+ TH2: 2x+3 < 0 => x < -3/2

cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16

=> (2x+3)(-5-2-1) = 16

=> 2x+3 = -2

=> 2x = -5

=> x = -5/2 (t/m)

/8(2x+3/ = 16

/2x+3/=2

2x+3=2 hoặc 2x+3=-2

2x=-1 hoặc 2x=-5

x=-1/2 hoặc x=-5/2

bạn trả lời nhé

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-6x^2\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+6x^2=5\)

\(\Leftrightarrow6x^2-6x+6=0\)

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0;\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0;\forall x\)

\(\Rightarrow\) Phương trình vô nghiệm

Ta có: \(\left(2x-1\right)^3-2x\left(4x^2-3x\right)=5\)

\(\Leftrightarrow8x^3-6x^2+12x-1-8x^3+6x^2=5\)

\(\Leftrightarrow12x=6\)

hay \(x=\dfrac{1}{2}\)

a: \(\left(\sqrt{3}\right)^x=243\)

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

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

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

=>\(x< -\dfrac{3}{2}\)

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

b) 3x - 6 - (8x + 4) - (10x + 15) = 50

=> 3x - 6 - 8x - 4 - 10x - 15  = 50

=> (3x - 8x - 10x)  =  6+ 4 + 15 + 50

=> -15x = 75 => x = 75 : (-15) = -5

c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số  có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)

+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3

+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1

Vậy x = 5/3 hoặc x = 1

a) (n-1)ⁿ⁺¹¹-(n-1)ⁿ=0

(n-1)ⁿ(n-1)¹¹-(n-1)ⁿ=0

(n-1)ⁿ[(n-1)¹¹-1]=0

(n-1)ⁿ=0 hoặc (n-1)¹¹-1=0

n-1=0   hoặc  (n-1)¹¹   =1

n=1      hoặc  n-1         =1

n=1      hoặc   n          =2

\(x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)

\(x^3-25x-x^3-8=3\)

\(-25x-8=3\)

\(-25x=3+8=11\)

\(x=\frac{-11}{25}\)

\(2x^2-10x-3x-2x^2=26\)

\(2x^2-13x-2x^2=26\)

\(-13x=26\)

\(x=-2\)

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow\left(2x^2-10x\right)-\left(3x+2x^2\right)=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

3(x-2)-4(2x+1)-5(2x+3)=50

3x-6-8x-4-10x-13=50

-15x-23=50

-15x=50+23

-15x=73

x=73:(-15)

x=\(-\frac{73}{15}\)

15x = 75 suy ra x = 5

xin lỗi em, chị nhầm, để chị làm lại nha:

3(x-2)-4(2x+1)-5(2x+3)=50

3x-6-8x-4-10x-15=50

-15x-25=50

-15x=50+25

-15x=75

x=75:(-15)

x=-5

em xem lần chị làm lại này nha. chúc em học tốt