BT1: Viết các biểu thức sau dưới dạng bình phương1 tổng(hiệu):
a, \(x^2+12x+36\)
b, \(x^2-x+\dfrac{1}{4}\)
BT2: Rút gọn biểu thức:
a, \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
b, \(\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
c, \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2\)
BT3: Tìm x:
a, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
b, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
c, \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2+1\right)=0\)
BT4: Tìm GTNN của:
\(A=x^2-4x+7\)
\(B=2x^2-6x\)
Giúp mình với mình cần gấp!!! Cảm ơn các bạn nhiều nha!!!
1)
a) \(x^2+12x+36=\left(x+6\right)^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Tick nha
3)
a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8\)
\(\Leftrightarrow-2x=7\)
\(\Rightarrow x=\dfrac{-7}{2}\)
b) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2\right)-5x+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-10x^2+2x+4x^2-5x+1=28\)
\(\Leftrightarrow0-3x^2+23x+28=28\)
\(\Leftrightarrow-3x^2+23x=0\)
\(\Leftrightarrow-x\left(3x-23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-23=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{3}\end{matrix}\right.\)
c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6-2x^4-2x^2-1=0\)
\(\Leftrightarrow-5x^4+x^2-2=0\)
Đặt \(-5t^2+t-2=0\)
\(\Delta=1^2-4\left(-5\right)\left(-2\right)=-39< 0\)
\(\Rightarrow PTVN\)
2)
a) \(\left(x^2-1\right)^3-\left(x^4+x^2-1\right)\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left[\left(x^2-1\right)^2-x^4-x^2+1\right]\)
b) \(\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)=\left(x^2+3\right)\left[x^4-3x^2+9-\left(x^2+3\right)^2\right]\)c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-3x^2+39x+6\)