x^2-6y^2=1
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\(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)
\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\) (ĐK: \(x\ne0,x\ne-1\))
\(\Leftrightarrow\dfrac{360\left(x+1\right)}{x\left(x+1\right)}-\dfrac{400x}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow360\left(x+1\right)-400x=x\left(x+1\right)\)
\(\Leftrightarrow360x+360-400x=x^2+x\)
\(\Leftrightarrow-40x+360=x^2+x\)
\(\Leftrightarrow x^2+40x+x-360=0\)
\(\Leftrightarrow x^2+41x-360=0\)
\(\Rightarrow\Delta=41^2-4\cdot1\cdot\left(-360\right)=3121>0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-41+\sqrt{3121}}{2\cdot1}\approx7\left(tm\right)\\x_2=\dfrac{-41-\sqrt{3121}}{2\cdot1}\approx-48\left(tm\right)\end{matrix}\right.\)
\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\)
Điều kiện: \(x\ne0;x\ne-1\)
PT \(\Leftrightarrow\dfrac{360\left(x+1\right)-400x}{x\left(x+1\right)}=1\)
\(\Rightarrow-40x+360=x\left(x+1\right)\)
\(\Leftrightarrow-40x+360=x^2+x\)
\(\Leftrightarrow x^2+41x-360=0\)
\(\Leftrightarrow x^2+2.\dfrac{41}{2}.x+\dfrac{1681}{4}=\dfrac{3121}{4}\)
\(\Leftrightarrow\left(x+\dfrac{41}{2}\right)^2=\left(\dfrac{\sqrt{3121}}{2}\right)^2\)
\(\Leftrightarrow x+\dfrac{41}{2}=\dfrac{\sqrt{3121}}{2}\) hoặc \(x+\dfrac{41}{2}=-\dfrac{\sqrt{3121}}{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\) hoặc \(x=-\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\)
Vậy...
(x-1)^3-x(x-2)^2+1
= x^3-3x^2+3x-1-x(x^2-4x+4)+1
= x^3-3x^2+3x-1- x^3+4x^2-4x+1
= x^2-x
= x(x-1)
HỌC TỐT!
@Zịt_siu_lừi
\(=x^3-3x^2+3x-1-x\left(x^2-4x+4\right)+1\)
\(=x^3-3x^2+3x-1-x^3+4x^2-4x+1\)
\(=x^2-x\)
có thể điền vào x là : 10 -7 là ước của 48 - 9.
b. 46 + 2 chia hết cho \(4^2\)
học tốt nha bạn
\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x^2-1\right)\left(x-3\right)^2=\left(x-1\right)\left(x+1\right)\left(x-3\right)^2\)
\(a,A=\left|2-4x\right|-6\ge-6\\ A_{min}=-6\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,x^2+1\ge1\Leftrightarrow B=1-\dfrac{4}{x^2+1}\ge1-\dfrac{4}{1}=-3\\ B_{min}=-3\Leftrightarrow x=0\)