Cho a.b.c.d \(\in\)R. CMR:
a) \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)
b) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
Những câu hỏi liên quan
Cho a,b,c,d∈R.CMR a2+b2≥2ab(1) Áp dụng cm các bđt sau:
a)\(a^4+b^4+c^4+d^4\ge4abcd\)
b)\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)
c) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
Ta có: \(a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
a) \(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2\ge4abcd\)
b) \(a^2+1\ge2a,b^2+1\ge2b,c^2+1\ge2c\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)
c) \(a^2+4\ge4a,b^2+4\ge4b,c^2+4\ge4c,d^2+4\ge4d\)
\(\Rightarrow\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
a) \(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2=2\left[\left(ab\right)^2+\left(cd\right)^2\right]\ge2\cdot2abcd=4abcd\)
b) \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a\cdot2b\cdot2c=8abc\)
c) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge4a\cdot4b\cdot4c\cdot4d=256abcd\)
Bài 3. Cho \(a,b,c\in R\). Chứng minh các bất đẳng thức sau:
\(a,\frac{a^2+3}{\sqrt{a^2+2}}>2\)
\(b,\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a^2+b^2\right)\) \(\left(ab>0\right)\)
\(c,\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
a)đpcm<=>(a2+3)2>4(a2+2)<=>(a2+1)2>0(lđ)
b)đpcm<=>\(a^4+b^4\ge ab\left(a^2+b^2\right)\)
Theo AM-GM\(\left\{{}\begin{matrix}a^4+b^4+b^4+b^4\ge4a^3b\\b^4+a^4+a^4+a^4\ge4b^3a\end{matrix}\right.\)
=>đpcm. Dấu bằng xảy ra khi a=b
c)AM-GM:\(VT\ge256\left|abcd\right|\ge256abcd\)
Dấu bằng xảy ra khi hai số bằng 2, hai số còn lại bằng -2 hoặc cả 4 số bằng 2 hoặc cả 4 số bằng -2
Rút gọn :
a,Aleft(3+1right)left(3^2+1right)left(3^4+1right)...left(3^{64}+1right)
b,B-1^2+2^2-3^2+4^2-...-99^2+100^2
c,C-1^2+2^2-3^2+4^2-...+left(-1right)^ncdot n^2
d,D3cdotleft(2^2+1right)left(2^4+1right)...left(2^{64}+1right)+1
e,Eleft(a+b+cright)^2+left(a+b-cright)^2-2left(a+bright)^2
g,Gleft(a+b+c+dright)^2+left(a+b-c-dright)^2+left(a+c-b-dright)^2+left(a+d-b-cright)^2
h,Hleft(a+b+cright)^3-left(b+c-aright)^3-left(a+c-bright)^3+left(a+b-cright)^3
i,Ileft(a+bright)^3+left(b+cright)^3+left(c...
Đọc tiếp
Rút gọn :
\(a,A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ b,B=-1^2+2^2-3^2+4^2-...-99^2+100^2\\ c,C=-1^2+2^2-3^2+4^2-...+\left(-1\right)^n\cdot n^2\\ d,D=3\cdot\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ e,E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\\ g,G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\\ h,H=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3+\left(a+b-c\right)^3\\ i,I=\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(c+b\right)\left(c+a\right)\)
Mọi người ơi, giúp mk vs, đc câu nào hay câu ấy ! Help me!!!!!!!!!!!!!!!!!!
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
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e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c\right)\left(a+b-c\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)
g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)
\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)
\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)
\(=2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)
\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)
\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
bn đăng nhiều quá nên mk làm câu nào hay câu đó nha
mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .
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Xem thêm câu trả lời
Tìm a,b,c biết
a, \(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2< =0\)
b,\(\left(a-7\right)^2+\left(3b+2\right)^2+\left(4c-5\right)^6< =0\)
c,\(\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+19\right)^6< =0\)
d,\(\left(7b-3\right)^4+\left(21a-6\right)^4+\left(18c+5\right)^6< =0\)
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
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Rút gọn
a) Aleft(3x+1right)^2-2left(3x+1right)left(3x+5right)+left(5x+5right)^2
b) Bleft(3+1right)left(3^2+1right)left(3^4+1right)left(3^8+1right)left(3^{18}+1right)left(3^{32}+1right)
c) Cleft(a+b-cright)^2+left(a-b+cright)^2-2left(b-cright)^2
d) Dleft(a+b+cright)^2+left(a-b-cright)^2+left(b-c-aright)^2+left(c-b-aright)^2
e)Eleft(a+b+c+dright)^2+left(a+b-c-dright)^2+left(a+c-b-dright)^2+left(a+d-b-cright)^2
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Rút gọn
a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
b) \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{18}+1\right)\left(3^{32}+1\right)\)
c) \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
d) \(D=\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-b-a\right)^2\)
e)\(E=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)
\(A=\left(-2x-4\right)^2\)
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A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2
= [(3x + 1)-(5x + 5)]2
= (3x + 1 - 5x - 5)2
= [(-2x) - 4]2
B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (38 - 1)(38 + 1)(316 +1)(332 + 1)
= (316 - 1)316 +1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
vì 2B = 364 - 1
=> B = \(\dfrac{3^{64}-1}{2}\)
C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)
= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2
= 2a2
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Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
a) A left(a+b+cright)^3+left(a-b+cright)^3-6aleft(b+cright)^2b) B left(2+1right)left(2^2+1right)left(2^4+1right)left(2^8+1right)left(2^{16}+1right)c) C 5left(2x-1right)^2+4left(x-1right)left(x+3right)-2left(5-3xright)^2d) D left(9x-1right)^2+left(1-5xright)^2+2left(9x-1right)left(1-5xright)e) E left(2a^2+2a+1right)left(2a^2-2a+1right)-left(2a^2+1right)^2
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a) A= \(\left(a+b+c\right)^3+\left(a-b+c\right)^3-6a\left(b+c\right)^2\)
b) B= \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
c) C= \(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
d) D= \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
e) E= \(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
Rút gọn rồi tính giá trị biểu thức :
a) Aleft(x+3right)^2+left(x-3right).left(x+3right)-2.left(x+2right).left(x-4right); với x -frac{1}{2}
b) Bleft(3x+4right)^2-left(x-4right).left(x+4right)-10x; với x -frac{1}{10}
c) Cleft(x+1right)^2-left(2x-1right)^2+3.left(x-2right).left(x+2right); với x 1
d) Dleft(x-3right).left(x+3right)+left(x-2right)^2-2x.left(x-4right); với x -1
Đọc tiếp
Rút gọn rồi tính giá trị biểu thức :
a) \(A=\left(x+3\right)^2+\left(x-3\right).\left(x+3\right)-2.\left(x+2\right).\left(x-4\right)\); với x = \(-\frac{1}{2}\)
b) \(B=\left(3x+4\right)^2-\left(x-4\right).\left(x+4\right)-10x\); với x = \(-\frac{1}{10}\)
c) \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3.\left(x-2\right).\left(x+2\right)\); với x = 1
d) \(D=\left(x-3\right).\left(x+3\right)+\left(x-2\right)^2-2x.\left(x-4\right)\); với x = -1
Cho \(A=\left\{x\in R|\left(x+1\right)^2+\left(x-1\right)^2=10\right\};B=\left\{x\in R|\left(x+1\right)^4+\left(x-1\right)^4=82\right\}\)Tìm tập X sao cho A\(\cup\)X=B.