b: =>x-3=12

hay x=15

a. 4.(x+41) = 7

x + 41 = 7 : 4 = 1,75

x = 1,75 - 41 = -39,25

b. 4.(x-3) = 7² - 1¹⁰ = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x = 12 + 3 = 15

a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)

a/ \(4.\left(x+41\right)=400\)

\(x+41=\dfrac{400}{4}=100\)

\(\Rightarrow x=59\)

b/ \(4.\left(x-3\right)=7^2-1^{10}\)

\(4.\left(x-3\right)=49-1\)

\(4.\left(x-3\right)=48\)

\(x-3=\dfrac{48}{3}=16\)

\(\Rightarrow x=19\)

#AEZn8

a ) 4 . ( x + 41 ) = 400

             x + 41    = 400 : 4

             x + 41    = 100

             x             = 100 - 41

             x             = 59

b ) 4 . ( x - 3 ) = 7²- 1¹⁰

      4 . ( x - 3 ) = 49 - 1

      4 . ( x - 3 ) = 48

             x - 3    = 48 : 4

             x - 3     = 12

             x           = 12 + 3

             x           = 15 

a: =>x-8=27

hay x=35

a. 25 + 3(x-8) = 106

3(x-8) = 106 - 25 = 81

x -8 = 81 : 3 = 27

x = 27 +8 = 35

b. 3²(x +4) - 5² = 5.5²

9(x+4) - 25 = 5.25 = 125

9(x+4) = 125 + 25 = 150(Đề sai)

3.(x-8) = 106-25

3.(x-8) = 81

x-8 = 81:3

x-8 = 27

x=35

\(a,25+3\left(x-8\right)=106\\ 3\left(x-8\right)=81\\ x-8=27\\ x=35\)

3² . (x + 4 ) - 5² = 5 . 2²

9  . ( x + 4 ) - 25 = 20

9 . ( x + 4 )        = 20 + 25

9 . ( x + 4 )        = 45

     ( x + 4 )       = 45 : 9

      x  + 4          = 5

      x                 = 5 - 4

      x                 = 1

a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1

a: =>x+41=7/4

hay x=-157/4

a) 4(x+41)=7
x+41=7/4
x=7/4-41
x=157/4
b) có gì đó sai sai-.-

Lời giải:

a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$

$\Leftrightarrow -4x.6=8$

$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$

b.

$9x^5-72x^2=0$

$\Leftrightarrow 9x^2(x^3-8)=0$

$\Leftrightarrow x^2=0$ hoặc $x^3=8$

$\Leftrightarrow x=0$ hoặc $x=2$

c.

$5x^4-8x^2-4=0$

$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$

$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$

$\Leftrightarrow (5x^2+2)(x^2-2)=0$

$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)

$\Leftrightarrow x=\pm \sqrt{2}$

d.

PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$

$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$

$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$

$\Leftrightarrow x+2=0$ hoặc $x+1=0$

$\Leftrightarrow x=-2$ hoặc $x=-1$

a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)

\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)

\(\Leftrightarrow-24x=8\)

hay \(x=-\dfrac{1}{3}\)

b: Ta có: \(9x^5-72x^2=0\)

\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)

\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)

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

c: Ta có: \(5x^4-8x^2-4=0\)

\(\Leftrightarrow5x^4-10x^2+2x^2-4=0\)

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

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)

\(\Rightarrow2x=-4\Rightarrow x=-2\)

b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)