`a, x^2-10x+25=0`

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

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

`<=>x-5=0`

`<=>x=5`

__

`x^2 -8x+16=0`

`<=> x^2 - 2.x.4+4^2=0`

`<=>(x-4)^2=0`

`<=>x-4=0`

`<=>x=4`

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`x^2-49=0`

`<=>x^2 - 7^2=0`

`<=>(x-7)(x+7)=0`

`<=>x-7=0` hoặc `x+7=0`

`<=> x=7` hoặc `x=-7`

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`4x^2-25=0`

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

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

`<=>2x-5=0` hoặc `2x+5=0`

`<=> 2x=5` hoặc `2x=-5`

`<=>x=5/2` hoặc `x=-5/2`

a: =>(x-5)^2=0

=>x-5=0

=>x=5

b: =>(x-4)^2=0

=>x-4=0

=>x=4

c: =>(x-7)(x+7)=0

=>x-7=0 hoặc x+7=0

=>x=7 hoặc x=-7

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

=>2x-5=0 hoặc 2x+5=0

=>x=5/2 hoặc x=-5/2

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

a) x² + 10x + 25 - 4x² - 20x = 0

<=> 3x² + 10x - 25 = 0

<=> (x + 5)(3x - 5) = 0 <=> \(\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)

Vậy S = \(\left\{-5;\frac{5}{3}\right\}\)

b. (4x - 5)² - 2(4x - 5)(4x + 5) = 0

<=> (4x - 5)[(4x - 5) - 2(4x + 5)] = 0

<=> (4x - 5)(4x - 5 - 8x - 10) = 0

<=> (4x - 5)(-4x - 15) = 0 <=> \(\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}\)

Vậy S = \(\left\{-\frac{15}{4};\frac{5}{4}\right\}\)

Tham khảo bài này :

(3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy x = -1/3 hoặc x = -5

\(a,x^2+10x+25-4x\left(x+5\right)=0.\)

\(\Leftrightarrow\left(x+5\right)^2-4x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(5-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\5-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

\(b,\left(4x-5\right)^2-2\left(16x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-5\right)^2-2\left(4x+5\right)\left(4x-5\right)=0\)

\(\Leftrightarrow-\left(4x-5\right)\left(4x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-5=0\\4x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}}\)

a. 4x²-25=0

=> (2x)²-5²=0

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

=> 2x-5=0    hoặc 2x+5=0

=> 2x=5        hoặc 2x=-5

=> x=5:2       hoặc x=-5:2

=> x=2,5        hoặc x=-2,5

b. (x-1)(4x²-49)=0

=> (x-1)[(2x)²-7²   ]=0

=> (x-1)(2x-7)(2x+7)=0

=> x-1=0     hoặc 2x-7=0           hoặc 2x+7=0

=> x=1         hoặc 2x=7               hoặc 2x=-7

=> x=1         hoặc x=7:2=3,5       hoặc x=-7:2=-3,5

a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow\sqrt{x-1}=1\)

hay x=2

c: Ta có: \(\sqrt{1-2x^2}=x-1\)

\(\Leftrightarrow1-2x^2=x^2-2x+1\)

\(\Leftrightarrow-3x^2+2x=0\)

\(\Leftrightarrow-x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

x = 1,25

Ta có : 

\(4x^2-25=0\)

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=-5\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-5}{2}\end{cases}}\)

Vậy phương trình có nghiệm \(x=\frac{5}{2}\) hoặc \(x=\frac{-5}{2}\)

Chúc bạn học tốt ~ 

\(4x^2-25=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-5}{2}\end{cases}}\)

a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )  

<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0

<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0

<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0

<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)

Vậy x = { \(\frac{-1}{3};-5\)} 

b) x² + 10.x + 25 - 4.x . ( x + 5 ) = 0 

<=> ( x + 5 )² -4.x . (x + 5 ) = 0

<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0

<=> ( x + 5 ) . ( 5 - 3.x )  = 0

<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{3};-5\right\}\)

c) (4.x - 5 )² - 2. ( 16.x² -25 ) = 0 

<=> ( 4.x-5)² -2 .( 4.x-5) .( 4.x + 5 ) = 0

<=> (  4.x -5 )² - ( 8.x+ 10 ) . ( 4.x -5 ) = 0

<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0

<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0

<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)

d) ( 4.x + 3 )² = 4. ( x² - 2.x + 1 ) 

<=> 16.x² + 24.x + 9 - 4.x²  + 8.x - 4 = 0

<=> 12.x² + 32.x + 5 =0 

<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0 

<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

e) x² -11.x + 28 = 0

<=> x² -4.x  - 7.x + 28 = 0

<=> ( x - 7 ) . ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy x = { 4 ; 7 } 

f ) 3.x.3 - 3.x² - 6.x = 0

<=> 3.x. ( x² -x - 2 ) = 0 

<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

        \([x=0\)                \([x=0\)

( Lưu ý :Lưu ý này không cần ghi vào vở :  Chị nối 2 ý đó làm 1 nha cj ! ) 

Vậy x = { 2 ; -1 ; 0 } 

cac ban giup minh nha

đề câu a khó hiểu thế

b/ 4x²-25=0

 <=>(2x)²-5²=0

<=>(2x-5)(2x+5)=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+5=0\\2x-5=0\end{array}\right.\)

\(\Leftrightarrow x=\pm\frac{5}{2}\)

c/ x³-4x²+4x=0

<=>x(x²-4x+4)=0

<=>x(x-2)²=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)