\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

\(ĐK:x\le-3;x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

`#040911`

`(x + 5)^3 = (2x)^3`

`\Rightarrow x + 5 = 2x`

`\Rightarrow x + 5 - 2x = 0`

`\Rightarrow 5 + (x - 2x) = 0`

`\Rightarrow 5 - x = 0`

`\Rightarrow x = 5 - 0`

`\Rightarrow x = 5`

Vậy, `x= 5.`

Cảm ơn bạn nha

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

\(\Rightarrow2x\left(x-3\right)=x\left(x-3\right)\)

\(\Rightarrow2x=x\)

\(\Rightarrow x=0\)

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

\(\left(x-3\right)=x^2-3x:2x\)

Ta có: \(2x\left(x-3\right)=x^2-3x\)

\(\Leftrightarrow2x\left(x-3\right)-x\left(x-3\right)=0\)

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

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

\(3\left(x+2\right)^2-5^2=2.5^2\)

\(\Rightarrow3\left(x+2\right)^2=2.5^2+5^2\)

\(\Rightarrow3\left(x+2\right)^2=5^2\left(2+1\right)\)

\(\Rightarrow3\left(x+2\right)^2=5^2.3\)

\(\Rightarrow\left(x+2\right)^2=5^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\) \(\Rightarrow x=3\left(x\inℕ\right)\)

3(x + 2)² - 5² = 2.5²

3(x + 2)² - 25 = 50

3(x + 2)² = 50 + 25

3(x + 2)² = 75

(x + 2)² = 75 : 3

(x + 2)² = 25

x + 2 = 5 hoặc x + 2 = -5

*) x + 2 = 5

x = 5 - 2

x = 3 (nhận)

*) x + 2 = -5

x = -5 - 2

x = -7 (loại)

Vậy x = 3

\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)

\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)

\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(5^x-2-3^2=2^4-0\)

\(5^x-2-3^2=2^4\)

\(5^x-2-9=16\)

\(5^x-2=16+9\)

\(5^x-2=25\)

\(5^x=25+2\)

\(5^x=27\)

Bởi vì 27 không phân tích được 1 số có số mũ là 2

\(\Rightarrow\) Không tồn tại x

\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)

\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}-9=16\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow x-2=25:5\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(5^{x-2}-3^2=2^4-\left(2^8\cdot2^4-2^{10}\cdot2^2\right)\)

\(\Rightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}=16+9\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow5^{x-2}=5^2\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=2+2\)

\(\Rightarrow x=4\)

Vậy: x=4

a) 5x + 7x + x = 169

12x + x = 169

13x = 169

x = 169 : 13

x = 13

b) x + x + x - 2 = 13

3x - 2 = 13

3x = 13 + 2

3x = 15

x = 15 : 3

x = 5

3.(⅓x - ¼)² = ⅓ 

=> (\(\dfrac{1}{3x}\)- \(\dfrac{1}{4}\) )² = \(\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{-1}{3}\\\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}=\dfrac{-1}{12}\\\dfrac{1}{3x}=\dfrac{7}{12}\end{matrix}\right.\)        => \(\left[{}\begin{matrix}x=-4\\x=\dfrac{12}{21}=\dfrac{4}{7}\end{matrix}\right.\)

Vậy, tập nghiệm x thỏa mãn là S=\(\left\{-4;\dfrac{4}{7}\right\}\)