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\(A=\left(x-4\right)^2-\left(x+4\right)^2-16\left(x-2\right)\)
\(=x^2-8x+16-x^2-8x-16-16x+32\)
\(=-32x+32\)
Biểu thức phụ thuộc vào giá trị của biến
a/ \(x^3=5x-12\Leftrightarrow x^3-5x+12=0\Leftrightarrow\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(4x+12\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)-3x\left(x+3\right)+4\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+4\right)=0\)
*) x + 3 = 0 <=> x = -3
S = {-3}
b/ có ng giải
c/ \(\left(2x^2-5x+3\right)^2=\left(x^2+x-2\right)^2\Leftrightarrow\left(2x^2-5x+3\right)^2-\left(x^2+x-2\right)^2=0\)
\(\Leftrightarrow\left(2x^2-5x+3-x^2-x+2\right)\left(2x^2-5x+3+x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-6x+5\right)\left(3x^2-4x-1\right)=0\)
\(\Leftrightarrow\left[\left(x^2-x\right)-\left(5x+5\right)\right]\left(3x^2-4x+1\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-5\left(x-1\right)\right]\left(3x^2-4x+1\right)=0\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(3x^2-4x+1\right)=0\)
*) x- 5 = 0 <=> x = 5
*) x- 1 = 0 <=> x = 1
S={1;5}
d/ \(x^3-x^2=4\left(x-1\right)^2\Leftrightarrow x^3-x^2-4\left(x-1\right)^2=x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-5x^2+8x-4=\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2=0\)
*) x - 1 = 0 <=> x = -1
*) (x - 2)^2 = 0 <=> x = 2
S = {-1;2}
\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)
Vô nghiệm
\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)
Đặt \(u=x^2-x\)
Phương trình trở thành \(u^2-4u+4=0\)
\(\Leftrightarrow\left(u-2\right)^2=0\)
\(\Leftrightarrow u-2=0\)
\(\Rightarrow x^2-x=2\)
\(\Rightarrow x^2-x-2=0\)
Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+3}{2}=2\\x=\frac{1-3}{2}=-1\end{cases}}\)
Đặt \(2x+1=w\)
Phương trình trở thành \(w^2-w=2\)
\(\Rightarrow\orbr{\begin{cases}w=2\\w=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=2\\2x+1=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
a)\(2+\frac{3}{x-5}=1\)
\(\Rightarrow\frac{3}{x-5}=-1\)
\(\Rightarrow3=-x+5\)
\(\Leftrightarrow x+3=5\)
\(\Rightarrow x=2\)
\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2-2x-3-x^2+3x-10\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases};x-13=0}\)
\(\Leftrightarrow x=1;x=2\)hoặc \(x=13\)
\(b,\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)+6\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+2x-x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)=0\)
Lại do \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+5\frac{3}{4}\ge5\frac{3}{4}>0\)
\(\Rightarrow\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)