\(\left(a1^2+a2^2\right)\cdot\left(b1^2+b2^2\right)>\left(a1b1+a2b2\right)\)
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TT
3
PT
9 tháng 12 2021
Câu 4: Cho bảng tính như hình bên:
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Hãy điền kết quả vào bảng sau:
Công thức tại ô D1 | Kết quả |
=SUM(A1:C3,1) |
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=AVERAGE(A2:C2) |
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=MIN(A2:C2,4) |
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=MAX(A3:C3) |
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=AVERAGE(A3:C3) + MAX(A2:C2) |
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22 tháng 12 2021
\(P=2+\dfrac{2}{b}+a+\dfrac{a}{b}+2+\dfrac{2}{a}+b+\dfrac{b}{a}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\left(\dfrac{3}{2a}+\dfrac{3}{2b}\right)+4\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{a.\dfrac{1}{2a}}+2\sqrt{b.\dfrac{1}{2b}}+2\sqrt{\dfrac{3}{2a}.\dfrac{3}{2b}}+4=6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\)
Ta lại có: \(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\left(BĐT.Cauchy\right)\Rightarrow2\left(a^2+b^2\right)\ge4ab\Rightarrow\sqrt{ab}\le\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow P\ge6+2\sqrt{2}+\dfrac{3}{\sqrt{ab}}\ge6+2\sqrt{2}+\dfrac{3}{\dfrac{\sqrt{2}}{2}}=6+5\sqrt{2}\)
\(minP=6+5\sqrt{2}\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)
4 tháng 1
`#3107.101107`
`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`
`= -9 + [-54 \div (-256 + 7) * 4]`
`= -9 + [-54 \div (-249) * 4]`
`= -9 + (18/83 * 4)`
`= -9 + 72/83`
`= -675/83`
______
`31 * (-18) + 31 * (-81) - 31`
`= 31 * (-18 - 81 - 1)`
`= 31 * (-100)`
`= -3100`
___
`(-12) * 47 + (-12) * 52 + (-12)`
`= (-12) * (47 + 52 + 1)`
`= (-12) * 100`
`= -1200`
___
`13 * (23 + 22) - 3 * (17 + 28)`
`= 13 * 45 - 3 * 45`
`= 45 * (13 - 3)`
`= 45 * 10`
`= 450`
____
`-48 + 48 * (-78) + 48 * (-21)`
`= 48 * (-1 - 78 - 21)`
`= 48 * (-100)`
`= -4800`
TN
17 tháng 6 2019
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
19 tháng 6 2019
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
DL
19 tháng 6 2019
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
go de ma cung go sai
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2
day la 1 dinh li noi tieng
<=> phan h 2 hai ve sau do bot cac hang tu giong nhau
<=> a^2d^2+b^2c^2>=2abcd
<=> (ad+bc)^2>=0 (luon dung )
=> DPCM
d ở đâu