3. Tìm x  biết

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

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

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

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

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

\(\Rightarrow\left(3x+1\right)^2-\left(5x-2\right)^2=0\)

\(\Rightarrow\left(3x+1-5x+2\right).\left(3x+1+5x-2\right)=0\)

\(\Rightarrow\left(-2x+3\right).\left(8x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-2x+3=0\\8x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=-3\\8x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{8}\end{cases}}\)

\(4x^2-25+\left(2x+5\right).\left(2x+7\right)=0\)

\(\Rightarrow\left(4x^2-25\right)+\left(2x+5\right).\left(2x+7\right)=0\)

\(\Rightarrow\left(2x-5\right).\left(2x+5\right)+\left(2x+5\right).\left(2x+7\right)=0\)

\(\Rightarrow\left(2x+5\right).\left(2x-5+2x+7\right)=0\)

\(\Rightarrow\left(2x+5\right).\left(4x+2\right)=0\)

\(\Rightarrow2.\left(2x+5\right).\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+5\right).\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+5=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-5\\2x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-1}{2}\end{cases}}\)

\(x^3+4x^2+4x=0\)

\(\Rightarrow x.\left(x^2+4x+4\right)=0\)

\(\Rightarrow x.\left(x+2\right)^2=0\)

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

\(x^2-10x+9=0\)

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

\(\Rightarrow x.\left(x-1\right)-9.\left(x-1\right)=0\)

\(\Rightarrow\left(x-9\right).\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-9=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

\(-3x^2+2x+1=0\)

\(\Rightarrow-3x^2+3x-x+1=0\)

\(\Rightarrow-3x.\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(-3x-1\right).\left(x-1\right)=0\)

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

2. Phân tích đa thức sau thành nhân tử

\(x^4+64\)

\(=x^4+4x^3+8x^2-4x^3-16x^2-32x+8x^2-32x+64\)

\(=x^2.\left(x^2+4x+8\right)-4x.\left(x^2+4x+8\right)+8.\left(x^2+4x+8\right)\)

\(=\left(x^2-4x+8\right).\left(x^2+4x+8\right)\)

\(x^4-x^2+1\) (Bạn xem lại đề nhé.)

 nếu ko có vào tcn nhé

\(A=-2x^2+6x+9\)

\(A=-\left(2x^2-6x-9\right)\)

\(A=-[\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\frac{3\sqrt{2}}{2}+\frac{9}{2}-\frac{27}{2}]\)

\(A=-\left(x\sqrt{2}-\frac{9}{2}\right)^2+\frac{27}{2}\le\frac{27}{2}\)

Dấu bằng xảy ra \(\Leftrightarrow x\sqrt{2}-\frac{9}{2}=0\)

\(\Leftrightarrow x\sqrt{2}=\frac{9}{2}\)

\(\Leftrightarrow x=\frac{9\sqrt{2}}{4}\)

Bài 1 :

A = x² + 2x + 2

A = 2 + ( x² + 2^x )

A = 2 + X² + 2

A = 2 + 2x²

A = 2¹ + 2x²

A = 2¹ + 2² x x

A = 2² - 2¹

A = 2¹ x

A = 2^X2

A = x = - \(\frac{1}{2}\)

A = x : 1¹ = - \(\frac{1}{2}\)

A = x² : - \(\frac{1}{2}\)

A = - \(\frac{1Z}{2}\)

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

\(=3x^2+3xy+3y^2-9xy=3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2=27\)(vì x - y = 3)

M = ab - (ab + bc + ca) + (a + b + c) - 1

= -(bc + ca) + (a + b) + (c - 1) 

= -c(b + a) + (a + b) + (c - 1) 

= (a + b)(-c + 1) + (c - 1)

Thay c = 1 vào M

=> M = (a + b)(-1 + 1) + (1 - 1) = 0

Vậy M = 0 

Kết quả = 0 nha. Bn thay c =1 r phá ngoặc là tính ra