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-.-

\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)

\(\Leftrightarrow x-\dfrac{1}{3}=3\)

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

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

\(\Leftrightarrow x=\dfrac{3}{20}\)

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: Ta có: \(x\left(2-x\right)+x^2+x=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)

\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)

\(\Leftrightarrow9x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{9}\)

hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)

c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)

\(\Leftrightarrow x^2-4x+5=0\)

\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý)

a: Ta có: \(71-\left(x+33\right)=26\)

\(\Leftrightarrow x+33=45\)

hay x=12

b: Ta có: \(\left(x+73\right)-26=76\)

\(\Leftrightarrow x+73=102\)

hay x=29

c: Ta có: \(45-\left(x+9\right)=6\)

\(\Leftrightarrow x+9=39\)

hay x=30

a) 71 - ( 33 + x ) = 26

(33 + x ) = 71 -26

33 + x = 45

x        = 45 - 33 = 12

b) ( x + 73 ) - 26 = 76

    ( x + 73 )        = 102

    x                     = 102 - 73 = 29

c) 45 - ( x + 9 ) = 6

 ( x + 9 )           = 45 - 6 = 39

 x                      = 39 - 9 = 30

d) 89 - ( 73 - x ) = 20 

 73 - x  = 89 - 20

73 - x = 69

x          = 73 - 69 = 4

e) 4(x+41) = 400

   x + 41 = 400 : 4 = 100

   x = 100 - 41 = 59

f) 11(x-9 ) = 77

     x - 9 = 77 : 11 = 7

    x = 7 + 9 = 16

g) x + 7 = 25 + 13 = 38

    x = 38 - 7 = 31

h) x + 4 = 198 - 120 = 78

   x = 78 - 4 = 74

i) x - 9 = 350 : 5 = 70

   x = 70 + 9 = 79

j) 2x - 49 = 5 . 9 = 45

   2x = 45 + 49 = 94

   x = 94 : 2 = 47

k) 25 + 3( x - 8 ) = 106

    3(x-8 ) = 106 - 25 = 81

   x - 8 = 81 : 3 = 27 

x = 27 +8= 35

l) 9( x + 4 ) - 25 = 20

   9( x + 4 ) = 20 + 25 = 45

     x + 4 =45 : 9 = 5

   x = 5 - 4 = 1

m) 200 - ( 2x + 6 ) = 64

     2x + 6 = 200 - 64 = 136

   2x = 136 - 6 = 130

x = 130 : 2 = 65

n) 2(x-51) = 16 + 20 = 36

   x - 51 = 36 : 2 = 13

  x = 13 + 51 = 64

o) 450 : ( x - 19 ) = 50

  x - 19 = 450 : 50 = 9

x = 19 + 9 = 28

p)4(x-3) = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x + 15

q) 156 - ( x+61 ) = 82

x + 61 = 156 - 82 = 74

x = 74 - 61 = 13

r) ( x - 35 ) - 120 = 0

x - 35 = 120

x = 120 + 35 = 155

a) \(\sqrt{x^2}=7\)⇒\(\left(\sqrt{x^2}\right)^2=49\)⇒x=7 hoặc -7

b) \(\sqrt{x^2}=8\)⇒\(\left(\sqrt{x^2}\right)=64\)⇒x=8 hoặc -8

c) \(\sqrt{4x^2}=6\)⇒\(\left(\sqrt{\left(2x\right)^2}\right)^2=36\)⇒x=3 hoặc -3

d) \(\sqrt{9x^2}=\left|-12\right|\)⇒\(\left(\sqrt{\left(3x\right)^2}\right)^2=144\)⇒x=12 hoặc -12

a. \(\sqrt{x^2}=7\)

<=> \(|x|=7\)

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

b. \(\sqrt{x^2}=8\)

<=> \(|x|=8\)

<=> \(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

c. \(\sqrt{4x^2}=6\)

<=> \(|2x|=6\)

<=> \(\left[{}\begin{matrix}2x=6\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

a: x=12

b: x=-19

c)x=-25

d)x=30

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

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

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

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

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

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

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

a)=> x² + 6x + 9 - x² + 4x = 39 => 10x = 30 => x = 3
b) => x(x - 9) + 2(x -9) = 0 => (x+2)(x-9) = 0 
+)th1: x + 2 = 0 => x = -2
+)th2: x - 9 =0 => x = 9