c: \(\Leftrightarrow x-3=0\)

hay x=3

c: 

hay x=3

mấy chế nhanh giúp mik vs

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 } 

\(x^4+\left(x+1\right)\left(5x^2-6x-6\right)=0\)

\(\Leftrightarrow x^4+5x^3-x^2-12x-6=0\)

\(\Leftrightarrow x^4-x^3+6x^3-x^2-6x^2+6x^2\)

\(-6x-6x-6=0\)

\(\Leftrightarrow\left(x^4-x^3-x^2\right)+\left(6x^3-6x^2-6x\right)+\)

\(\left(6x^2-6x-6\right)=0\)

\(\Leftrightarrow x^2\left(x^2-x-1\right)+6x\left(x^2-x-1\right)+\)

\(6\left(x^2-x-1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+6\right)\left(x^2-x-1\right)=0\)

\(TH1:x^2+6x+6=0\)

Ta có: \(\Delta=6^2-4.6=12\sqrt{\Delta}=\sqrt{12}\)

pt có 2 nghiệm:

\(x_1=\frac{-6+\sqrt{12}}{2}=-3+\sqrt{3}\)

\(x_2=\frac{-6-\sqrt{12}}{2}=-3-\sqrt{3}\)

\(TH2:x^2-x-1=0\)

Ta có: \(\Delta=1^2+4.1=5,\sqrt{\Delta}=\sqrt{5}\)

pt có 2 nghiệm:

\(x_1=\frac{1+\sqrt{5}}{2}\)và \(x_2=\frac{1-\sqrt{5}}{2}\)

Vậy pt có 4 nghiệm \(x_1=\frac{-6+\sqrt{12}}{2}=-3+\sqrt{3}\);\(x_2=\frac{-6-\sqrt{12}}{2}=-3-\sqrt{3}\);

\(x_3=\frac{1+\sqrt{5}}{2}\);\(x_4=\frac{1-\sqrt{5}}{2}\)

Làm tốt rồi nhưng mà lớp 8 chưa học cách giải pt bậc 2 \(\Delta\). Thì chúng ta có thể:

VD TH1: \(x^2+6x+6=0\)

<=> \(x^2+6x+9-9+6=0\)

<=> \(\left(x+3\right)^2=3\)

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

tương tự Th2.

\(PT< =>x^4+5x^3-6x^2-6x+5x^2-6x-6=0\)

\(< =>x^4+5x^3-x^2-12x-6=0\)

\(< =>\left(x^2-x-1\right)\left(x^2+6x+6\right)=0\)

<=>\(\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{cases}}\)hay \(\orbr{\begin{cases}x=-3+\sqrt{3}\\x=-3-\sqrt{3}\end{cases}}\)

Vậy \(S=\left\{\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2};-3+\sqrt{3};-3-\sqrt{3}\right\}\)