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Đặt \(P=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(P=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{ab}=1+\dfrac{a+b}{ab}+\dfrac{1}{ab}=1+\dfrac{2}{ab}\)
Với mọi a;b dương, ta có:
\(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow4ab\le1\Rightarrow ab\le\dfrac{1}{4}\Rightarrow\dfrac{1}{ab}\ge4\)
\(\Rightarrow P\ge1+2.4=9\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
a) \(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(=\left(3x^2-6x+3\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)
\(=-30\)
b) \(B=-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)
\(=-x\left(x^2+4x+4\right)+\left(4x^2+4x+1\right)+\left(x^3-3x^2+9x+3x^2-9x+27\right)-1\)
\(=27\)
a: Ta có: \(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-4x^2-12x-9-5+20x\)
\(=-30\)
b: Ta có: \(B=-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)
\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1\)
=27
\(x^2+y^2-2x+4y+8=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy GTNN là 3 khi x=1 và y=-2
=>Chọn B
1) \(x^2-5x-6=x^2-6x+x-6\)
\(=\left(x^2-6x\right)+\left(x-6\right)\)
\(=x\left(x-6\right)+\left(x-6\right)\)
\(=\left(x-6\right)\left(x+1\right)\)
2) \(2x^2-4xy+2y^2=2x^2-2xy-2xy+2y^2\)
\(=\left(2x^2-2xy\right)-\left(2xy-2y^2\right)\)
\(=2x\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-2y\right)\)
\(=\left(x-y\right).2\left(x-y\right)\)
\(=2\left(x-y\right)^2\)
HỌC TỐT NHA !!
\(x^2-6x+x-6\)
\(\left(x^2-6x\right)+\left(x-6\right)\)
\(x\left(x-6\right)+\left(x-6\right)\)
\(\left(x-6\right)\left(x+1\right)\)
\(x-6=0\) hoặc \(x+1=0\)
\(x=6\) hoặc \(x=-1\)
ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)
\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)
Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé
\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)
\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)
\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)
\(\Leftrightarrow4x+10=0\)
\(\Leftrightarrow4x=-10\)
\(\Leftrightarrow x=\frac{-5}{2}\)
a: s=v*t
b: Theo đề, ta có: x/10-x/12=1/4
=>x/60=1/4
=>x=15