a) \(x^2-4x=0\)

\(x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

b) \(4x^2-9=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)

c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

d) \(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-2\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

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

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

\(\left(x-3\right)\left(3x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)

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

\(x.\left(x-4\right)=0\)

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

\(4x^2-9=0\)

\(2^2x^2-9=0\)

\(\left(2x\right)^2-9=0\)

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

\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\cdot\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

\(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-\left(4x+18\right)=0\)

\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)

\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

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

\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)

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

\(\)

\(A=x^3+3xy^2-9+y\left(3x^2+y^2\right)\)

\(=x^3+3x^2y+3xy^2+y^3-9\)

\(=\left(x+y\right)^3-9\)

\(=\left(1,95+0,05\right)^3-9=2^3-9=-1\)

\(B=x^2+\frac{1}{2}x+\frac{1}{16}\)

\(=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)

\(=\left(x+\frac{1}{4}\right)^2\)

\(=\left(9,75+0,25\right)^2=10^2=100\)

\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

Vậy \(\orbr{\begin{cases}a=-2,b=2\\a=2,b=2\end{cases}}\)

Chúc bạn học tốt.

Ta có:

\(2x^2-5xy+3y^2\)

\(=2x^2-2xy-3xy+3y^2=2x\left(x-y\right)-3y\left(x-y\right)\)

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

\(x^3-7x-6=x^3+1-7x-7\)

\(=\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)

\(x^2+x=6\)

<=>  \(x^2+x-6=0\)

<=>   \(\left(x-2\right)\left(x+3\right)=0\)

tự lm tiếp

b)  \(6x^3+x^2=2x\)

<=>  \(6x^3+x^2-2x=0\)

<=>  \(x\left(6x^2+x-2\right)=0\)

<=>   \(x\left(2x-1\right)\left(3x+2\right)=0\)

tự giải ra

a/\(x^2+x=6\)

\(x\left(x+1\right)=6\)

=> TH1 :x =0

     TH2 : x+1 =0  nên x = ( -1 )

b/\(6x^3+x^2=2x\)

\(6x^3+x^2-2x=0\)

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

TH1 : 2x =0  nên x =0

TH2 : x-1 =0 nên x =1

TH2 : x+1 =0 nên x = (-1)

a)  \(4x^2-4x+3=4x^2-4x+1+2\)

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

=> ko phân tích thành nhân tử được

b)  \(9x^2+6x-8=9x^2+12x-6x-8\)

\(=3x\left(3x+4\right)-2\left(3x+4\right)=\left(3x-2\right)\left(3x+4\right)\)

c)  \(3x^2-8x+4=3x^2-6x-2x+4\)

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

a/\(4x^2-4x+3\)

\(=4x^2-1x-3x+3\)

\(=4x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(4x-3\right)\)

b/\(9x^2+6x-8\)

\(=\text{(3x - 2)(3x + 4)}\)

c/\(3x^2-8x+4\)

\(\text{ =(3x^2 - 6x) - (2x - 4) }\)

\(\text{= 3x(x - 2) - 2(x - 2)}\)

\(\text{= (3x - 2)(x - 2)}\)

  x^4 + 2x^3 - 4x^2 - 5x - 6 = 0 
<=>x^4 - 2x^3 + 4x^3 - 8x^2 + 4x^2 - 8x + 3x - 6 = 0 
<=> x^3(x - 2) + 4x^2(x - 2) + 4x(x - 2) + 3(x - 2) = 0 
<=>(x - 2)(x^3 + 4x^2 + 4x + 3) = 0 
<=>(x - 2)(x^3 + 3x^2 + x^2 + 3x + x + 3) = 0 
<=>(x - 2)[x^2(x + 3) + x(x + 3) + (x + 3)] = 0 
<=>(x - 2)(x + 3)(x^2 + x + 1) = 0 

Tìm ƯCLN của 1751 và 1957 
Nhập 1751/1957,máy hiện : 17/19 
=> ƯCLN (1751 ; 1957) = 1751/17 = 103 (số nguyên tố) 
Thử lại thì 2369 cũng chia hết cho 103 tức là 103 là 1 ước nguyên tố của 2369 
* Phân tích các hạng tử ra thừa số nguyên tố : 
1751^3 + 1957^3 + 2396^3 = (103.17)^3 + (103.19)^3 + (103.23)^3 
= 103^3.(19^3 + 17^3 + 23^3) = 103^3. 23939 = 103^3.37.647 

Dễ thấy 103, 37 và 647 là các số nguyên tố 
=> ước nguyên tố của 1751^3 + 1957^3 + 2369^3 là 103, 37, 647