Đặt \(a=24-x,b=x-25\)

Khi đó pt ban đầu trở thành :

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)

\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow30a^2+68ab+30b^2=0\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)

Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !

`@` `\text {Ans}`

`\downarrow`

`a,`

`(x - 2)(x - 3) =0`

`<=>`\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0+2\\x=0+3\end{matrix}\right.\)

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

Vậy, `S = {2; 3}`

`b,`

`x^2 - 5x = 0`

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

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

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

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

Vậy, `S = {0; 5}`

`c,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`d,`

`4x^2 - 25 = 0`

`<=> 4x^2 = 25`

`<=> x^2 = 25/4`

`<=> x^2 = (+-5/2)^2`

`<=> x = +-5/2`

Vậy,` S = {5/2; -5/2}.`

a: =>x-2=0 hoặc x-3=0

=>x=2 hoặc x=3

b: =>x(x-5)=0

=>x=0 hoặc x=5

c: =>(x-3)(x+3)=0

=>x=3 hoặc x=-3

d: =>(2x-5)(2x+5)=0

=>x=5/2 hoặc x=-5/2

a) -5 + |3x - 1| + 6 = |-4|

=> -5 + |3x - 1| + 6 = 4

=> 1 + |3x - 1| = 4

=> |3x - 1| = 4 - 1

=> |3x - 1| = 3

=> \(\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)

=> \(\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)

=>  \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy ...

d) |x + 1| + |x + 2| + |x + 3| = 4x

Ta có: |x + 1| \(\ge\)0  \(\forall\)x

          |x + 2| \(\ge\)0 \(\forall\)x

          |x + 3| \(\ge\)0 \(\forall\)x

=> |x + 1| + |x + 2| + |x + 3| \(\ge\)0 \(\forall\)x => 4x \(\ge\)0 \(\forall\) x=> x \(\ge\)0 \(\forall\)x

=> x + 1 + x + 2 + x + 3 = 4x

=> 3x + 6 = 4x

=> 6 = 4x - 3x

=> x = 6

Vậy...

b) (x - 1)² = (x - 1)⁴

=> (x - 1)² - (x - 1)⁴ = 0

=> (x - 1)² .[1 - (x - 1)² ] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\)

=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy x = {1; 2; 0}

\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)

\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)

\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)

\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)

a: =>(x+2-3)(x+2+3)=0

=>(x-1)(x+5)=0

=>x=1 hoặc x=-5

b: =>(x-1)^2=25

=>x-1=5 hoặc x-1=-5

=>x=-4 hoặc x=6

c: =>25x^2+10x+1-25x^2+9=30

=>10x+10=30

=>x+1=3

=>x=2

d: =>x^3-1-x(x^2-4)=5

=>x^3-1-x^3+4x=5

=>4x=6

=>x=3/2

Ta có: 25^{x + 1} . 125^x . 625^{x + 2} = (5²)⁵

=> (5²)^{x + 1} . (5³)^x  . (5⁴)^{x+ 1} = 5¹⁰

=> 5^{2x + 2} . 5^3x . 5^{4x + 8} = 5¹⁰

=> 2x + 2 + 3x + 4x + 8 = 10

=> 9x + 2 + 8 = 10

=> 9x = 10 - 2 - 8

=> 9x = 0

=> x = 0 : 9

=> x = 0

\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

a)\(-x^2\left(x^2-4\right)=-25\left(x^2-4\right)\)

\(\Leftrightarrow-x^2=-25\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm5\)

(2x + 1)³ = 125

(2x + 1)³ = 5³

=> 2x + 1 = 5

     2x       = 4

       x       = 2

\(\left(2x+1\right)^3=125=5^3\)

=>2x+1=5

2x=4

x=2

hok tốt

\((2x+1)^3=125\)

\(\Rightarrow(2x+1)^3=5^3\)

\(\Rightarrow2x+1=\pm5\)

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(x\in\left\{2;-3\right\}\)

1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)

=>3x-1/5=3/25 hoặc 3x-1/5=-3/25

=>3x=8/25 hoặc 3x=2/25

=>x=8/75 hoặc x=2/75

2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)

=>2x-1/3=2/9 hoặc 2x-1/3=-2/9

=>2x=5/9 hoặc 2x=1/9

=>x=5/18 hoặc x=1/18