a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

\(x^2+5x=\sqrt{37}\)

\(\Leftrightarrow4x^2+20x=4\sqrt{37}\)

\(\Leftrightarrow4x^2+20x+25=4\sqrt{37}+25\)

\(\Leftrightarrow\left(2x+5\right)^2=4\sqrt{37}+25\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=\sqrt{4\sqrt{37}+25}\\2x+5=-\sqrt{4\sqrt{37}+25}\end{matrix}\right.\Leftrightarrow x=\dfrac{\pm\sqrt{4\sqrt{37}+25}-5}{2}\)

\(x^2+5x=\sqrt{37}\)

\(\Leftrightarrow x\left(x+5\right)=\sqrt{37}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x+5=\sqrt{37}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x=-5+\sqrt{37}\end{matrix}\right.\)

Vậy \(x=\sqrt{37};x=-5+\sqrt{37}\).

a ) x = 25 28 b ) x = 5 12 .

c) x= 1             d) x = 17

a; \(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{7}\)

    \(x\)        = \(\dfrac{1}{7}\) + \(\dfrac{3}{4}\)

    \(x\)       =  \(\dfrac{4}{28}\) + \(\dfrac{21}{28}\)

    \(x\)       =  \(\dfrac{25}{28}\)

Vậy \(x=\dfrac{25}{28}\) 

b; - \(x\) + \(\dfrac{3}{5}\) + \(\dfrac{13}{20}\) = \(\dfrac{5}{6}\)

     (\(\dfrac{12}{20}\) + \(\dfrac{13}{20}\))  - \(x\) = \(\dfrac{5}{6}\)

    \(\dfrac{5}{4}\) - \(x\)  = \(\dfrac{5}{6}\) 

           \(x\) = \(\dfrac{5}{4}\) - \(\dfrac{5}{6}\)

           \(x\) = \(\dfrac{30}{24}\) -  \(\dfrac{20}{24}\)

            \(x\) = \(\dfrac{5}{12}\)

Vậy \(x=\dfrac{5}{4}\) 

\(25\cdot x^2=25\\ x^2=25\div25\\ x^2=1\\ x^2=1^2\\ x=1\\ 8^{x-3}=1\\ 8^{x-3}=8^0\\ x-3=0\\ x=0+3\\ x=3\)

`@` `\text {Ans}`

`\downarrow`

\(3^{x+6}=37\)

`=>`\(3^x\cdot3^6=37\)

`=>`\(3^x=37\div3^6\)

`=>` \(3^x=\dfrac{37}{729}\) 

Bạn xem lại đề.

\(25x^2=25\)

`=>`\(x^2=25\div25\)

`=>`\(x^2=1\)

`=> x=1`

\(8^{x-3}=1\)

`=>`\(8^x\div8^3=1\)

`=>`\(8^x=8^3\)

`=> x=3`

a) Tìm được x = 7 4 .       b) Tìm được x= 10 9 .  

a) \(\left[\left(x+32\right)-17\right].2=42\)

\(\left[\left(x+32\right)-17\right]=42:2\)

\(\left[\left(x+32\right)-17\right]=21\)

\(\left(x+32\right)=21+17\)

\(\left(x+32\right)=38\)

\(x=38-32\)

\(x=6\)

b) \(125+\left(145-x\right)=175\)

\(\left(145-x\right)=175-125\)

\(\left(145-x\right)=50\)

\(x=145-50\)

\(x=95\)

\(\text{Δ}=\left(4m+1\right)^2-8\left(m-4\right)\)

\(=16m^2+8m+1-8m+32\)

\(=16m^2+33>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

Ta có: \(\left|x_1-x_2\right|=17\)

\(\Leftrightarrow\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=17\)

\(\Leftrightarrow\sqrt{\left(4m+1\right)^2-4\cdot2\cdot\left(m-4\right)}=17\)

\(\Leftrightarrow\sqrt{16m^2+8m+1-8m+32}=17\)

\(\Leftrightarrow16m^2+33=289\)

=>m=4 hoặc m=-4

a: \(\Leftrightarrow x-3=-13\)

hay x=-10