\(x^2+4x-9y^2+4\)

\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)

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

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

\(x^2-9y^2-6y-1\)

\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)

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

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

Tham khảo nhé~

a/Ta có:x²+4x-9y²+4

=x²+4x+4-(3y)²

=(x+2)²-(3y)²

=(x+2-3y)(x+2+3y)

b/Ta có:x²-9y²-6xy-1

=x²-6xy-(3y)²-1

=(x-3y-1)(x-3y+1)

\(2x^2=x\)

\(\Rightarrow2x^2-x=0\)

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

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

Vậy \(x=0\)hoặc \(x=\frac{1}{2}\)

\(x^3=x^5\)

\(\Rightarrow x^5-x^3=0\)

\(x^3.\left(x^2-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=1\)

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

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

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

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

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

Vậy \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\) hoặc \(x=-2\)

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

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

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

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

Vậy \(x=\frac{3}{2}\)hoặc \(x=-2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

\(S\left\{0;\frac{1}{2}\right\}\)

\(d)x^3-x^5=0\Leftrightarrow x^3\left(1-x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{1}\end{cases}}\)

\(S=\left\{0;\pm\sqrt{1}\right\}\)

các câu sau tương tự nha bn

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

   \(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                 \(=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow A=\frac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)

TK nha!!

1)3x(x-2)=7(x-2)

<=>3x(x-2)-7(x-2)=0

<=>(x-2)(3x-7)=0

x-2=0=>x=2

3x-7=0=>x=7/3

cn lại lm tg tự

10)\(x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}\)

\(\left(\frac{x}{2}+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\frac{x}{4}+3\right)=0\)

\(\Rightarrow\frac{5x}{2}-3x^2+15-18x+3x^2+36x-\frac{x}{2}-6=0\)

\(\Rightarrow\frac{5x}{2}-\frac{x}{2}+15-6-\left(18x-36x\right)=0\)

\(\Rightarrow2x+9+18x=0\)

\(\Rightarrow20x=-9\)

\(\Rightarrow x=-\frac{9}{20}\)

lkcs

nha bn

LK + CS =HP

Có đúng ko bạn???

nếu đúng thì kb và k cho mk nhé!