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bài 5 nhé:
a) (a+1)2>=4a
<=>a2+2a+1>=4a
<=>a2-2a+1.>=0
<=>(a-1)2>=0 (luôn đúng)
vậy......
b) áp dụng bất dẳng thức cô si cho 2 số dương 1 và a ta có:
a+1>=\(2\sqrt{a}\)
tương tự ta có:
b+1>=\(2\sqrt{b}\)
c+1>=\(2\sqrt{c}\)
nhân vế với vế ta có:
(a+1)(b+1)(c+1)>=\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}\)
<=>(a+1)(b+1)(c+1)>=\(8\sqrt{abc}\)
<=>(a+)(b+1)(c+1)>=8 (vì abc=1)
vậy....
\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)
Câu 5:
\(a+b=1\Rightarrow a=1-b\)
\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)
\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)
Câu 7:
\(a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)
\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
5.
Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)
\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=a^2+b^2-ab\)
\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)
\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)
6.
Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)
Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)
Mặt khác với mọi a;b ta có:
\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)
Từ đó:
\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)
\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)
\(N_{max}=2\) khi \(a=b=1\)
\(1.a,\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(b,\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ad-bc\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow-\left(ad-bc\right)^2\le0\left(luôn-đúng\right)\)
\(dấu"='\) \(xảy\) \(ra\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(c2:x+y=2\Rightarrow\left(x+y\right)^2=4\)
\(\Rightarrow\left(x+y\right)^2+\left(x-y\right)^2\ge4\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2\ge4\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge4\Leftrightarrow x^2+y^2\ge2\)
\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=y=1\)
Câu 1:
a)Ta có (ac+bd)2+(ad-bc)2=(ac)2+2abcd+(bd)2+(ad)2-2abcd+(bc)2
=(ac)2+(bd)2+(ad)2+(bc)2
=a2(c2+d2)+b2(c2+d2)
=(a2+b2)(c2+d2) (đpcm)
b)Ta có (ac+bd)2 = (ac)2+2abcd+(bd)2
Lại có (a2+b2)(c2+d2) = (ac)2+(bd)2+(ad)2+(bc)2
Ta có (ac+bd)2 ≤ (a2+b2)(c2+d2)
<=>(a2+b2)(c2+d2) - (ac+bd)2 ≥ 0
<=>(ac)2+(bd)2+(ad)2+(bc)2-[(ac)2+2abcd+(bd)2]
<=>(ad)2 - 2abcd +(bc)2 ≥ 0
<=>(ad-bc)2 ≥ 0 (Luôn đúng) => đpcm
Câu 2:
Áp dụng BĐT Bunhiacôpxki, ta có (x+ y)2 ≤ (x2 + y2)(12 + 12) => 4 ≤ 2.S => 2 ≤ S
Dấu ''='' xảy ra <=> x=y=1
Vậy Min S=2 <=> x=y=1
1)chứng minh cái j ???
2)\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+2abcd+a^2d^2-2abcd+b^2c^2\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b)Ta có:
\(\left(ab+cd\right)^2\le\left(a^2+c^2\right)\left(b^2+d^2\right)\)
\(\Leftrightarrow a^2b^2+c^2d^2+2abcd\le a^2b^2+a^2d^2+b^2c^2+c^2d^2\)
\(\Leftrightarrow a^2d^2+b^2c^2-2abcd\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\)(Đpcm)
c)Áp dụng Bđt Bunhiacopxki ta có:
\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2=2^2=4\)
\(\Rightarrow2\left(x^2+y^2\right)\ge4\)
\(\Rightarrow x^2+y^2\ge2\)\(\Rightarrow S\ge2\)
Dấu = khi \(x=y=1\)
C1
Giả sử căn 7 là số hữu tỉ Vậy căn 7 bằng a/b. Suy ra 7 bằng a bình / b bình. Suy ra a bình bằng 7b bình Suy ra a chia hết cho 7 Gọi a bằng 7k suy ra a bình bằng 7b bình Suy ra (2k) bình bằng 2b bình suy ra 4k bình bằng 2b bình suy ra 2k bình bằng b bình Suy ra ƯCLN(a,b)=2 Trái với đề bài =>căn 7 là số vô tỉ
Câu 2a
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2-\left(a^2c^2+b^2d^2+a^2d^2+b^2c^2\right)=0\)
\(\Leftrightarrow0=0\)( đpcm )
Câu 2b
\(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow2abcd\le b^2c^2+a^2d^2\)
\(\Leftrightarrow0\le b^2c^2-2abcd+a^2d^2\)
\(\Leftrightarrow0\le\left(bc-ad\right)^2\)( đpcm )
Câu 4a
\(\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( đpcm )
Câu 4c
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow3a+5b\ge2\sqrt{3a.5b}=2\sqrt{15ab}\)
\(\Rightarrow12\ge2\sqrt{15ab}\)
\(\Rightarrow6\ge\sqrt{15ab}\)
\(\Rightarrow6^2\ge15ab\)
\(\Rightarrow36\ge15ab\)
\(\Rightarrow ab\le\frac{12}{5}\)
\(\Leftrightarrow P\le\frac{12}{5}\)
Vậy GTLN của \(P=\frac{12}{5}\)