Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(4x^2-4\)
\(=4\left(x^2-1\right)\)
\(=4\left(x-1\right)\left(x+1\right)\)
XXét tứ giác AMDN có ^AMD=^MAN=^AND=90∞
⇒AMDN là hình chữ nhật
hcn AMDN có AD là phân giác góc A
⇒AMDN là hình vuông(dấu hiệu 3)
Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x-9=0
<=> -2x=9
<=> \(x=\frac{-9}{2}\left(tmđk\right)\)
3x(x-1)=1-x
<=> 3x(x-1) +x-1=0
<=> (x-1)(3x+1)=0
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}}\)
Vậy...
\(\left(3x-1\right)^2.\left(x+5\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(x^2-4x-1=0\)
\(\left(x^2-2\cdot x\cdot2+4\right)-5=0\)
\(\left(x-2\right)^2=\left(\sqrt{5}\right)^2\)
\(\Rightarrow x-2=\pm\sqrt{5}\)
Tự giải tiếp nha ...
\(x^2-4x-1=0\)
\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\)
pt có 2 nghiệm
\(x_1=\frac{-4-\sqrt{20}}{2}=-2-\sqrt{5}\)
\(x_2=\frac{-4+\sqrt{20}}{2}=-2+\sqrt{5}\)