a: x=12

b: x=-19

c)x=-25

d)x=30

1: x=7+5=12

2: =>-29-x=-10

=>x+29=10

=>x=-19

3: =>x+7=-18

=>x=-25

4: =>x+19-11=0

=>x+8=0

=>x=-8

1,a,\(\left|x+2\right|=x+3\Leftrightarrow\orbr{\begin{cases}x+2=x+3\\x+2=-x-3\end{cases}}\)

                                      \(\Leftrightarrow\orbr{\begin{cases}x-x=3-2\\x+x=-3-2\end{cases}\Leftrightarrow\orbr{\begin{cases}0=1\left(voly\right)\\x=\frac{-5}{2}\end{cases}}}\)

b, \(|x-2|=2-x\Leftrightarrow\orbr{\begin{cases}x-2=2-x\\x-2=x-2\end{cases}}\)

                                    \(\Leftrightarrow\orbr{\begin{cases}x+x=2+2\\x-x=-2+2\end{cases}\Rightarrow x=2}\)

c,\(\left|2x-1\right|=3\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

                             \(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

d,\(\left|x-12\right|=x\Leftrightarrow\orbr{\begin{cases}x-12=x\\x-12=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}0=12\left(voly\right)\\x=6\end{cases}}}\)

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

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) \(1=\left(2x+0,5\right)^{600}\)

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

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

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

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

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+2)+(x+5)+(x+8)+(x+11)+(x+14)=75

=>(x+x+x+x+x)+(2+5+8+11+14)=75

=>Xx5+40=75

=>Xx5=75-40

=>Xx5=35

=>x=35:5

=>x=7

\(\dfrac{x-1}{101}-\dfrac{x-13}{103}=\dfrac{x-17}{107}-\dfrac{x-19}{109}\)

\(\Rightarrow\left(\dfrac{x-11}{101}+1\right)-\left(\dfrac{x-13}{103}+1\right)=\left(\dfrac{x-17}{107}+1\right)-\left(\dfrac{x-19}{109}+1\right)\)

\(\Rightarrow\dfrac{x+90}{101}-\dfrac{x+90}{103}=\dfrac{x+90}{107}-\dfrac{x+90}{109}\)

\(\Rightarrow\dfrac{x+90}{101}-\dfrac{x+90}{103}-\dfrac{x+90}{107}+\dfrac{x+90}{109}=0\)

\(\Rightarrow\left(x+90\right)\left(\dfrac{1}{101}-\dfrac{1}{103}-\dfrac{1}{107}+\dfrac{1}{109}\right)=0\)

Vì \(\dfrac{1}{101}-\dfrac{1}{103}-\dfrac{1}{107}+\dfrac{1}{109}\ne0\)

Nên \(x+90=0\Rightarrow x=-90\)

xy - x + 2y = 3

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

=> x(y-1) + 2(y-1) = 5

=> (x+2)(y+1) = 5

=> x + 2 và y + 1 \(\in\)Ư(5) = {-1;5;-5;1}

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