a) x = 4

b) x = 3

c) x = 2

d) x = 1

e) x = 3

f) x = 2

g) x = 4

h) x = 3

2:

1: =>36x+14x=69+81=150

=>50x=150

=>x=3

2: 3^x=81

=>3^x=3^4

=>x=4

3: 3(2x+1)^2=75

=>(2x+1)^2=25

=>2x+1=5 hoặc 2x+1=-5

=>x=-3 hoặc x=2

1:

1: \(\dfrac{13\cdot17^4+4\cdot17^4}{17^3}-\dfrac{14\cdot3^3-14\cdot3^2}{9}\)

\(=\dfrac{17^4\cdot\left(13+4\right)}{17^3}-\dfrac{14\cdot3^2\left(3-1\right)}{9}\)

\(=17\cdot17-14\cdot2\)

=289-28

=261

2:

\(2^3\cdot5^2-\left[131-\left(23-2^3\right)^2\right]\)

\(=8\cdot25-131+\left(-1\right)^2\)

=69+1

=70

\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)

\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)

\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)

\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)

dấu chấm là gì vậy

dấu chấm là dấu x trong cấp trên

a) x = 4                

b) x = 3     

c) x = 2                

d) x = 1. 

a: 49x^2-25=0

=>(7x-5)(7x+5)=0

=>7x-5=0 hoặc 7x+5=0

=>x=5/7 hoặc x=-5/7

b: Đề thiếu vế phải rồi bạn

c: (3x-2)^2-9(x+4)(x-4)=2

=>9x^2-12x+4-9(x^2-16)=2

=>9x^2-12x+4-9x^2+144=2

=>-12x+148=2

=>-12x=-146

=>x=146/12=73/6

d: x^3-6x^2+12x-8=0

=>(x-2)^3=0

=>x-2=0

=>x=2

e: x^3-9x^2+27x-27=0

=>(x-3)^3=0

=>x-3=0

=>x=3

a) \(-25+49x^2=0\)

\(\Leftrightarrow49x^2-25=0\)

\(\Leftrightarrow\left(7x\right)^2-5^2=0\)

\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)

b) \(16x^2-25\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)

\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)

c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)

\(\Leftrightarrow-84x-140=2\)

\(\Leftrightarrow-84x=142\)

\(\Leftrightarrow x=-\dfrac{142}{84}\)

\(\Leftrightarrow x=-\dfrac{71}{42}\)

d) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

e) \(-27+27x-9x^2+x^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Mình sửa lại câu c một chút nha:

c: (3x-2)^2-9(x+4)(x+4)=2

=>(3x-2)^2-9(x+4)^2=2

=>(3x-2)^2-(3x+12)^2=2

=>(3x-2-3x-12)(3x-2+3x+12)=2

=>-14*(6x+10)=2

=>6x+10=-1/7

=>6x=-71/7

=>x=-71/42

a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

=>-x^2(x-2)-(x-2)(x+2)=0

=>(x-2)(-x^2-x-2)=0

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)