All questions
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Nguyễn Thị Linh 06/03/2019 at 23:32
No, The letter math
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FacuFeri 06/03/2019 at 05:13
We have : \(x^2-180x+8102=\left(x^2-180x+8100\right)+2=\left(x-90\right)^2+2\ge2\forall x\left(1\right)\)
Applying the Bunhiacopxki inequality , we have :
\(\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le\left(1+1\right)\left(x-89+91-x\right)\)
\(\Rightarrow\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le2.2=4\)
\(\Rightarrow\sqrt{x-89}+\sqrt{91-x}\le2\left(2\right)\)
Because \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)
So ( 1 ) ; ( 2 ) ; we have : \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102=2\)
Equal sign occurs \(\Leftrightarrow\sqrt{x-89}=\sqrt{91-x};x-90=0\left(3\right)\)
We have : \(\sqrt{x-89}=\sqrt{91-x}\) \(\Leftrightarrow x-89=91-x\Leftrightarrow x=90\left(4\right)\)
( 3 ) ; ( 4 ) \(\Rightarrow x=90\) is the result of the equation
Lê Anh Duy selected this answer. -
Huy Toàn 8A (TL) 06/03/2019 at 13:14
Đk : \(89\le x\le91\)
Applying the Bunhiacopxki inequality
We have : VT \(=1.\sqrt{x-89}+1.\sqrt{91-x}\le\sqrt{\left(1+1\right)\left(x-89+91-x\right)=2}\)
=> VT \(\le2\)
And VP = \(x^2-2.x.90+90^2+2=\left(x-90\right)^2+2\ge2\)
=> VP \(\ge2\ge\) VT
The sign "=" occurs when and only when:
\(\left\{{}\begin{matrix}\sqrt{x-89}=\sqrt{91-x}\\x-90=0\end{matrix}\right.\) => \(x=90\)(satisfy)
The answer is 90
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FacuFeri 05/03/2019 at 05:10
Tran Anh ??? Le Quoc Tran Anh ???
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Huy Toàn 8A (TL) 03/03/2019 at 13:37
El matodora : The Spam question
You should write the real question in English
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Alone 03/03/2019 at 15:25
The speed limit in this zone in feet per second is:
Quoc Tran Anh Le selected this answer.
88:60 x 15==22(feet) -
nguyễn kim tuyen 14/04/2019 at 02:40
22 feet
Quoc Tran Anh Le Coordinator
02/03/2019 at 13:19
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Huy Toàn 8A (TL) 02/03/2019 at 13:29
C1: We have : 3x + 4x + 5x = x (3 + 4 + 5)
Replace x = 10 <=> 10 ( 3 + 4 + 5) = 10 . 12 = 120
C2 : 3x + 4x + 5x <=> 3 . 10 + 4 . 10 + 5 . 10 = 30 + 40 + 50 = 120
Quoc Tran Anh Le selected this answer.
Quoc Tran Anh Le Coordinator
28/02/2019 at 15:14
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Huy Toàn 8A (TL) 01/03/2019 at 04:01
We have
\(\dfrac{2}{8}.\dfrac{5}{7}=256.78125=20000000\)
\(=>1000000000:\left(\dfrac{2}{8}.\dfrac{5}{7}\right)=1000000000:20000000=50\)
The answer is 50
Quoc Tran Anh Le selected this answer. -
Thanh Hải Đặng 16/03/2019 at 05:26
1 000 000 000 =109=29.59
\(\Rightarrow\)1000000000 / (28x57)=2*52=50
The answer ís 50
Quoc Tran Anh Le Coordinator
28/02/2019 at 15:14
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Kaka Lot 02/03/2019 at 13:33
This is two