x+2/x-2 -x-2/x+2=-16/4x2
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Những câu hỏi liên quan
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2021
\(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
\(4x^2-4x+1-\left(y^2+8y+16\right)=\left(2x-1\right)^2-\left(y+4\right)^2\)
\(=\left(2x-1-y-4\right)\left(2x-1+y+4\right)=\left(2x-y-5\right)\left(2x+y+3\right)\)
14 tháng 7 2021
Ta có: \(A=\left(x-2\right)\left(x^4+2x^3+4x^2+8x+16\right)\)
\(=x^4+2x^3+4x^2+8x+16\)
\(=3^4+2\cdot3^3+4\cdot3^2+8\cdot3+16\)
\(=81+54+36+24+16\)
\(=211\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
11 tháng 8 2023
\(A=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\\ =\left[\left(5x-2\right)+\left(6x+1\right)\right].\left[\left(5x-2\right)-\left(6x+1\right)\right]+11\left(x^2-4\right)-48+32x\\ =-\left(11x-1\right)\left(x+3\right)+11x^2-44-48+32x\\ =-11x^2-32x+3+11x^2-44-48+32x\\ =-11x^2+11x^2-32x+32x+3-44-48=-89\)
Vậy biểu thức A không phụ thuộc vào giá trị của x
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
11 tháng 8 2023
\(B=4x\left(x-3\right)-\left(x-5\right)^2-3\left(x+1\right)^2+\left(2x+2\right)^2-\left(4x^2-5\right)\\ =4x^2-12x-\left(x^2-10x+25\right)-3\left(x^2+2x+1\right)+\left(4x^2+8x+4\right)-4x^2+5\\ =4x^2-x^2-3x^2+4x^2-4x^2-12x+10x-6x+8x+25-3+4+5\\ =31\)
Vậy giá trị biểu thức B không phụ thuộc biến x
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{-16}{4x2}\left(ĐK:x\ne\pm2;0\right)\)
\(< =>\frac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=-\frac{16}{8x}\)
\(< =>\frac{\left(x+2\right)\left(x-2\right)-\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2}=-\frac{2}{x}\)
\(< =>\frac{0}{\left(x+2\right)^2}=-\frac{2}{x}\)
\(< =>0=-2\left(x+2\right)^2\)
\(< =>2\left(x+2\right)^2=0< =>x+2=0\)
\(< =>x=-2\left(ktmđk\right)\)
Vậy phương trình trên vô nghiệm
\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=-\frac{16}{4x^2}\left(x\ne\pm2;0\right)\)
\(\frac{\left(x+2\right)^24x^2}{\left(x+2\right)\left(x-2\right)4x^2}-\frac{\left(x-2\right)^24x^2}{\left(x+2\right)\left(x-2\right)4x^2}=\frac{-16\left(x+2\right)\left(x-2\right)}{4x^2\left(x+2\right)\left(x-2\right)}\)
Khử mẫu và rút gọn ta đc : \(32x^3=-16x^2+64\Leftrightarrow32x^3+16x^2-64=0\)
Hooc ne vào là đẹp !