a) 𝑎𝑏̅.101
b) 𝑎𝑏̅.10001
c) 𝑎𝑎𝑎𝑎̅∶𝑎
d) 𝑎𝑎𝑎𝑎̅∶𝑎𝑎̅
e) 𝑎𝑎𝑎𝑎𝑎𝑎̅∶𝑎𝑎𝑎̅
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5ab = 3ab
5ab - 3ab = 0
3ab.(2ab - 1) = 0
⇔\orbr{3ab=02ab−1=0⇔\orbr{ab=0ab=12
a)65 . ( - 101 )
=65.(−100−1)=65.(−100−1)
=65(−100)−65.1=65(−100)−65.1
=−6500−65=−6500−65
=−6565
b)
1 2 3 4 5 6 7 8
1 2 3 4 \(|\)trái
\(|\)3 4phải
\(|\)4phải
D
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)
\(minA=1\Leftrightarrow x=10\)
\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)
\(minB=-201\Leftrightarrow x=-10\)
\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)
\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)
\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
b: ta có: \(B=2x^2+40x-1\)
\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)
\(=2\left(x+10\right)^2-201\ge-201\forall x\)
Dấu '=' xảy ra khi x=-10
a) 𝑎𝑏̅.101 = abab
b) 𝑎𝑏̅.10001 = ab00ab
c) 𝑎𝑎𝑎𝑎̅∶𝑎 = 1111
d) 𝑎𝑎𝑎𝑎̅∶𝑎𝑎̅ = 101
e) 𝑎𝑎𝑎𝑎𝑎𝑎̅∶𝑎𝑎𝑎̅ = 1001
cam on thuy tien nha