a/ Cho a>b . Chứng minh rằng : a+2 > b-3
b/ a;b thuộc số thực dương
Chứng minh rằng : \(\frac{1}{a}\)+\(\frac{1}{b}>=\frac{4}{a+b}\)
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a)a2+b2+c2+3=2(a+b+c)
=>a2+b2+c2+1+1+1-2a-2b-2c=0
=>(a2-2a+1)+(b2-2b+1)+(c2-2c+1)=0
=>(a-1)2+(b-1)2+(c-1)2=0
=>a-1=b-1=c-1=0 <=>a=b=c=1
-->Đpcm
b)(a+b+c)2=3(ab+ac+bc)
=>a2+b2+c2+2ab+2ac+2bc -3ab-3ac-3bc=0
=>a2+b2+c2-ab-ac-bc=0
=>2a2+2b2+2c2-2ab-2ac-2bc=0
=>(a2- 2ab+b2)+(b2-2bc+c2) + (c2-2ca+a2) = 0
=>(a-b)2+(b-c)2+(c-a)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
c)a2+b2+c2=ab+bc+ca
=>2(a2+b2+c2)=2(ab+bc+ca)
=>2a2+2b2+c2=2ab+2bc+2ca
=>2a2+2b2+c2-2ab-2bc-2ca=0
=>a2+a2+b2+b2+c2+c2-2ab-2bc-2ca=0
=>(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ca+c2)=0
=>(a-b)2+(b-c)2+(a-c)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
a. \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)
\(\Leftrightarrow2a^2+2b^2=a^2+b^2-2ab\)
\(\Leftrightarrow a^2+b^2=-2ab\)
\(\Leftrightarrow a^2+2ab+b^2=0\)
\(\Leftrightarrow\left(a+b\right)^2=0\)
\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\) (đpcm)
b. \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì \(\left(a-1\right)^2;\left(b-1\right)^2;\left(c-1\right)^2\ge0\)
\(\Rightarrow\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
\(\Leftrightarrow a-1=b-1=c-1=0\Leftrightarrow a=b=c=1\)
c. \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tương tự câu b ta có a = b = c
\(a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow ab+a+ab+b=ab+a+b+1\Leftrightarrow ab=1\left(dpcm\right)\)
Ta có: A= 2 + 22 + 23 + ... + 260= (2 +22) + (23+ 24) + ... + (259 + 260).
= 2 x (2 + 1) + 23 x (2 + 1) + ... + 259 x (2 + 1).
= 2 x 3 + 23 x 3 + ... + 259 x 3.
= 3 x ( 2 + 23 + ... + 259).
Vì A = 3 x ( 2 + 23 + ... + 259) nên A chia hết cho 3.
A= (2 +22 + 23) + (24 + 25 + 26) + ... + (258 + 259 + 260).
= 2 x (1 + 2 + 22) + 24 x (1 + 2 + 22) + ... + 258 x (1 + 2 + 22).
= 2 x 7 + 24 x 7 + ... + 258 x 7.
= 7 x ( 2 + 24 + ... + 258).
Vì A = 7 x ( 2 + 24 + ... + 258) nên A chia hết cho 7.
A= (2 +22 + 23 + 24) + (25 + 26 + 27 + 28) + ... + (257 + 258 + 259 + 260).
= 2 x (1 + 2 + 22 + 23) + 25 x (1 + 2 + 22 + 23) + ... + 257 x (1 + 2 + 22 + 23).
= 2 x 15 + 25 x 15 + ... + 257 x 15.
= 15 x ( 2 + 24 + ... + 258).
Vì A = 15 x ( 2 + 24 + ... + 258) nên A chia hết cho 15.
Ta có: B= 3 + 33 + 35 + ... + 31991= (3 + 33 + 35) + (37+ 39 + 311 ) + ... + (31987 + 31989 + 31991).
= 3 x (1 + 32 + 34) + 37 x (1 + 32 + 34) + ... + 31987 x (1 + 32 + 34).
= 3 x 91 + 37 x 91 + ... + 31987 x 91= 3 x 7 x 13 + 37 x 7 x 13 + ... + 31987 x 7 x 13.
= 13 x ( 3 x 7 + 37 x 7 + ... + 31987 x 7).
Vì B = 13 x ( 3 x 7 + 37 x 7 + ... + 31987 x 7) nên B chia hết cho 13.
B= (3 + 33 + 35 + 37) + ... + (31985 + 31987 + 31989 + 31991).
= 3 x (1 + 32 + 34 + 36) + ... + 31985 x (1 + 32 + 34 + 36).
= 3 x 820 + ... + 31985 x 820= 3 x 20 x 41 + ... + 31985 x 20 x 41.
= 41 x ( 3 x 20 + .. + 31985 x 20)
Vì B =41 x ( 3 x 20 + .. + 31985 x 20) nên B chia hết cho 41.
a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)
\(=3\times91+3^7\times91+...+3^{1987}\times91\)
\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)
\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)
Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.
b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)
\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)
\(=3\times820+...+3^{1985}\times820\)
\(=3\times20\times41+...+3^{1985}\times20\times41\)
\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)
Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.
1a)Xét a2 + 5 - 4a =a2 - 4a + 4+1=(a - 2)2+1\(\ge\)1 hay (a -2)2 + 1 > 0
\(\Rightarrow\)Đpcm
b)Xét 3(a2 + b2 + c2) -(a + b +c)2 =3a2 + 3b2 + 3c2 - a2 - b2 - c2 - 2ab - 2ac - 2bc
=2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc
=(a - b)2 + (a - c)2 + (b - c)2\(\ge\)0 (với mọi a,b,c)
\(\Rightarrow\)Đpcm
2)Xét A=\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+c+b\right)=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)
áp dụng cô-sy
\(\Rightarrow\)A\(\ge\)9
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)
a.
Bình phương 2 vế, BĐT cần chứng minh trở thành:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)
Ta có:
\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cộng lại:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
b.
\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)
Nên ta chỉ cần chứng minh:
\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)
\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)
Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)