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\(1)\left(\dfrac{1}{5}\right)^5\cdot5^5\\ =\left(\dfrac{1}{5}\cdot5\right)^5\\ =1^5\\ =1\\ 2)\left(\dfrac{2}{5}\right)^9\cdot5^9\\ =\left(\dfrac{2}{5}\cdot5\right)^9\\ =2^9\\ 3)\left(\dfrac{4}{9}\right)^3\cdot3^3\\ =\left(\dfrac{4}{9}\cdot3\right)^3\\ =\left(\dfrac{4}{3}\right)^3\\ 4)\left(\dfrac{3}{7}\right)^2\cdot\left(-7\right)^4\\ =\left(\dfrac{3}{7}\right)^2\cdot\left[\left(-7\right)^2\right]^2\\ =\left(\dfrac{3}{7}\right)^2\cdot49^2\\ =\left(\dfrac{3}{7}\cdot49\right)^2\\ =\left(3\cdot7\right)^2\\ =21^2\\ 5)\left(-11\right)^{12}\cdot\left(\dfrac{4}{11}\right)^6\\ =\left[\left(-11\right)^2\right]^6\cdot\left(\dfrac{4}{11}\right)^6\\ =121^6\cdot\left(\dfrac{4}{11}\right)^6\\ =\left(121\cdot\dfrac{4}{11}\right)^6\\ =\left(4\cdot11\right)^6\\ =44^6\\ 6)\left(-6\right)^8\cdot\left(\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-6\right)^7\cdot\left(\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-6\cdot\dfrac{5}{6}\right)^7\\ =\left(-6\right)\cdot\left(-5\right)^7\)
Kẻ H\(x\) // FG
Ta có : \(\widehat{xHI}\) = \(\widehat{JIH}\) = 450 (Hai góc so le trong)
\(\widehat{xHG}\) + \(\widehat{FGH}\) = 1800 (hai góc trong cùng phía)
⇒ \(\widehat{xHG}\) = 1800 - 1350 = 450
\(\widehat{IGH}\) = \(\widehat{xHG}\) + \(\widehat{xHI}\) = 450 + 450 = 900
Vậy HG vuông góc với HI
Kẻ H\(x\) // FG
Ta có : \(\widehat{xHI}\) = \(\widehat{JIH}\) = 450 (Hai góc so le trong)
\(\widehat{xHG}\) + \(\widehat{FGH}\) = 1800 (hai góc trong cùng phía)
⇒ \(\widehat{xHG}\) = 1800 - 1350 = 450
\(\widehat{IGH}\) = \(\widehat{xHG}\) + \(\widehat{xHI}\) = 450 + 450 = 900
Vậy HG vuông góc với HI
Bài 13:
\(1)A=x^2-x+1\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 2)B=x^2+x+1\\ =\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 3)C=x^2+2x+2\\ =\left(x^2+2x+1\right)+1\\ =\left(x+1\right)^2+1\ge1>0\forall x\)
\(4)A=x^2-5x+10\\ =\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{15}{4}\\ =\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\forall x\\ 5)B=x^2-8x+20\\ =\left(x^2-8x+16\right)+4\\ =\left(x-4\right)^2+4\ge4>0\forall x\\ 6)C=x^2-8x+17\\ =\left(x^2-8x+16\right)+1\\ =\left(x-4\right)^2+1\ge1>0\forall x\)
\(7)A=x^2-6x+10\\ =\left(x^2-6x+9\right)+1\\ =\left(x-3\right)^2+1\ge1>0\forall x\\ 8)B=9x^2-6x+2\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\ge1>0\forall x\\ 9)C=2x^2+8x+15\\ =\left(2x^2+8x+8\right)+7\\ =2\left(x^2+4x+4\right)+7\\ =2\left(x+2\right)^2+7\ge7>0\forall x\)
Gọi số đó có dạng \(\overline{ab}\)
Khi thêm số 0 vào giữa thì ta có số mới là: \(\overline{a0b}=100a+b\)
Mà số mới gấp 7 lần số cũ nên ta có:
\(\overline{a0b}=7\overline{ab}\\ 100a+b=7\left(10a+b\right)\\ 100a+b=70a+7b\\ 100a-70a=7b-b\\ a\left(100-70\right)=b\left(7-1\right)\\ 30a=6b\\ \dfrac{a}{b}=\dfrac{6}{30}=\dfrac{1}{5}\)
`=> a=1;b=5`
Vậy sso cần tìm là 15
Ta có BĐT Bunhiacopxki:
\(\left(1\cdot\sqrt{a}+1\cdot\sqrt{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\Leftrightarrow\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\) (*)
Dấu "=" xảy ra khi: \(\dfrac{\sqrt{a}}{1}=\dfrac{\sqrt{b}}{1}\Leftrightarrow a=b\)
a) \(2\le x\le4\)
Áp dụng bđt (*) ta có:
\(A=\sqrt{x-2}+\sqrt{4-x}\le\sqrt{2\left(x-2+4-x\right)}=2\)
Dấu "=" xảy ra khi: \(x-2=4-x\Leftrightarrow x=3\) (tm)
b) \(-2\le x\le6\)
Áp dụng bđt (*) ta có:
\(B=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{2\left(6-x+x+2\right)}=4\)
Dấu "=" xảy ra khi: \(6-x=x+2\Leftrightarrow x=2\left(tm\right)\)
c) \(0\le x\le2\)
\(C=\sqrt{x}+\sqrt{2-x}\le\sqrt{2\left(x+2-x\right)}=2\)
Dấu "=" xảy ra khi: \(x=2-x\Leftrightarrow x=1\left(tm\right)\)