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. `4x^2-20x+25=0`

`<=>(2x)^2-2.2x.5 +5^2=0`

`<=>(2x-5)^2=0`

`<=>2x-5=0`

`<=>x=5/2`

b. `(x-5)(x+5)-(x-3)^2=2(x-7)`

`<=>x^2-25-x^2+6x-9=2x-14`

`<=>6x-34=2x-14`

`<=>4x=20`

`<=>x=5`

\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)

\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)

\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)

a) Có: (2x)² - 2.2.5.x + 5² = 0

          ⇒ (2x - 5)² = 0 ⇒ 2x - 5 = 0

          ⇒ 2x = 5 ⇒ x = \(\dfrac{5}{2}\)

b) Có: x² - 25 - x² + 6x - 9 = 2x - 14

           ⇒ 6x - 36 = 2x - 14

           ⇒ 4x = 22

           ⇒ x = \(\dfrac{11}{2}\)

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

a) \(5x\left(x-2\right)+\left(2-x\right)=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)

\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)

a) \(1=\left(2x+0,5\right)^{600}\)

\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)

\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)

b) \(\left(x-0,125\right)^2=0,25\)

\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)

\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)

\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)

\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

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

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

`@` `\text {Ans}`

`\downarrow`

`a)`

`1 = (2x + 0,5)^600`

`=> (2x+0,5)^600 = (+-1)^600`

`=> \text {TH1: } 2x + 0,5 = 1`

`=> 2x = 1 - 0,5`

`=> 2x = 0,5`

`=> x = 0,5 \div 2`

`=> x = 0,25`

`\text {TH2: } 2x + 0,5 = -1`

`=> 2x = -1 - 0,5`

`=> 2x = -1,5`

`=> x = -1,5 \div 2`

`=> x = -0,75`

Vậy, `x \in {-0,75; 0,25}.`

`b)`

`(x - 0,125)^2 = 0,25`

`=> (x - 0,125)^2 = (+-0,5)^2`

`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)

Vậy, `x \in {-0,375; 0,625}.`

`c)`

`(x - 3)^11 = (x - 3)^41`

`=> (x - 3)^11 - (x - 3)^41 = 0`

`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`

`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy, `x \in {3; 4}.`

a, 1=(2x+0,5)⁶⁰⁰

=>1⁶⁰⁰=(2x+0,5)⁶⁰⁰

=>1=2x+0,5

=>2x+0,5=1

=> 2x= 1+0,5

=>2x= 1,5

=:>x= 1,5:2

=>x=0,75

b, (x-0,125)²=0,25

=>(x-0,125)²=(0,5)²

=>x-0,125=0,5

=>x=0,5 +0,125

=>x= 0,625

c, (x-3)¹¹=(x-3)⁴¹

(x-3)¹¹ -(x-3)⁴¹=0

(x-3)¹¹-(x-3)¹¹ .(x-3)³⁰=0

(x-3)¹¹.[1-(x-3)^{30 ]}=0

(x-3)¹¹.(4-x)³⁰=0

(x-3)¹¹=0 hoặc (4-x)³⁰=0

(x-3)¹¹=0          ( 4-x)³⁰=0

x-3=0                4-x=0

x=0+3                  x=4-0

x=3                     x=4

vậy xϵ{3,4}

a: x=12

b: x=-19

c)x=-25

d)x=30

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

\(a,\sqrt{x}=3\Leftrightarrow x=9\\ b,\sqrt{x}=6\Leftrightarrow x=36\\ c,\sqrt{x}=8\Leftrightarrow x=64\\ d,\sqrt{x}=12\Leftrightarrow x=144\\ e,2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\\ f,3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\\ g,\sqrt{x}=8\Leftrightarrow x=64\\ h,2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

a. \(\sqrt{x}=3\)

<=> \(\left(\sqrt{\sqrt{x}}\right)^2-\left(\sqrt{3}\right)^2=0\)

<=> \(\left(\sqrt{\sqrt{x}}-\sqrt{3}\right)\left(\sqrt{\sqrt{x}}+\sqrt{3}\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{\sqrt{x}}-\sqrt{3}=0\\\sqrt{\sqrt{x}}+\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\left(Vnghiêm\right)\end{matrix}\right.\)

Vậy nghiệm của PT là S = \(\left\{9\right\}\)

a) \(\sqrt{x}=3\Leftrightarrow x=9\)

b) \(\sqrt{x}=6\Leftrightarrow x=36\)

c) \(\sqrt{x}=8\Leftrightarrow x=64\)

d) \(\sqrt{x}=12\Leftrightarrow x=144\)

e) \(2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\)

f) \(3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)

g) \(\sqrt{x}=8\Leftrightarrow x=64\)

h) \(2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

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

b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

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