Cho a, b, c là các số dương thỏa mãn a + b + c = 1. Tìm Min:
\(A=\dfrac{\left(1+a\right)\cdot\left(1+b\right)\cdot\left(1+c\right)}{\left(1-a\right)\cdot\left(1-b\right)\cdot\left(1-c\right)}\)
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Bài 1:Tính
a, Adfrac{3}{4}cdotdfrac{8}{9}cdotdfrac{15}{16}cdot....cdotdfrac{9999}{10000}
b,Bleft(1-dfrac{1}{21}right)cdotleft(1-dfrac{1}{28}right)cdotleft(1-dfrac{1}{36}right)cdot....cdotleft(1-dfrac{1}{1326}right)
c,Cleft(1+dfrac{1}{1cdot3}right)cdotleft(1+dfrac{1}{2cdot4}right)cdotleft(1+dfrac{1}{3cdot5}right)cdot....cdotleft(1+dfrac{1}{99cdot101}right)
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Bài 1:Tính
a, A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot....\cdot\dfrac{9999}{10000}\)
b,B=\(\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot....\cdot\left(1-\dfrac{1}{1326}\right)\)
c,C=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
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Cho a,b,c là các số thực không âm thỏa mãn \(a^2\cdot b^2+c^2\cdot b^2+1\le3b\)
Tìm GTNN của biểu thức P=\(\frac{1}{\left(a+1\right)^2}+\frac{4\cdot b^2}{\left(1+2\cdot b\right)^2}+\frac{8}{\left(c+3\right)^2}\)
Cho a,b,c là các số thực dương có tổng bằng 1.CMR \(\left(a+\frac{1}{b}\right)\cdot\left(b+\frac{1}{c}\right)\cdot\left(c+\frac{1}{a}\right)\le\left(\frac{10}{3}\right)^3\)
cho a, b, c là 3 số thực khác 0, thỏa mãn
\(\frac{a+b-2017\cdot c}{c}=\frac{b+c-2017\cdot a}{a}=\frac{c+a-2017\cdot b}{b}\)
tính giá trị của biểu thức
B=\(\left(1+\frac{b}{a}\right)\cdot\left(1+\frac{a}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
Giải phương trình:
\(\dfrac{\left(x-a\right)\cdot\left(x-c\right)}{\left(b-a\right)\cdot\left(b-c\right)}+\dfrac{\left(x-b\right)\cdot\left(x-c\right)}{\left(a-b\right)\cdot\left(a-c\right)}=1\)
a, b. c là hằng số và khác nhau đôi một
\(\dfrac{ \left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-a\right)\left(x-c\right)+\left(x-b\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)+\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(x-a+x-b\right)}{\left(b-a\right)\left(b-c\right)-\left(b-a\right)\left(c-a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(b-c-c+a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(-2c+b+a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-\left(a+b\right)\right)}{\left(b-a\right)\left(-2c+\left(a+b\right)\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)2x}{\left(b-a\right)\left(-2c\right)}=1\)
\(\Leftrightarrow\dfrac{2x^2-2xc}{-2cb+2ac}=1\)
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a,b,c nguyên dương và a+b+c=1. chứng minh rằng \(\left(1+\dfrac{1}{a}\right)\cdot\left(1+\dfrac{1}{b}\right)\cdot\left(1+\dfrac{1}{c}\right)>=64\)
Lời giải:
Ta có:
\(\text{VT}=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\) (1)
Thay \(1=a+b+c\) kết hợp với bất đẳng thức AM-GM:
\((a+1)(b+1)(c+1)=(a+a+b+c)(b+a+b+c)(c+a+b+c)\)
\(=[(a+b)+(a+c)][(b+c)(b+a)][(c+a)+(c+b)]\)
\(\geq 2\sqrt{(a+b)(a+c)}.2\sqrt{(b+c)(b+a)}.2\sqrt{(c+a)(c+b)}\)
\(\Leftrightarrow (a+1)(b+1)(c+1)\geq 8(a+b)(b+c)(c+a)\)
Tiếp tục áp dụng AM-GM:
\((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\)
Suy ra \((a+1)(b+1)(c+1)\geq 64abc\) (2)
Từ (1);(2) ta có \(\text{VT}\geq 64\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)
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21 tháng 10 2017 lúc 15:04
Cho a,b,c là 3 số thực, thỏa mãn điều kiện:
a)\(\dfrac{a+b-c}{3\cdot c}=\dfrac{b+c-a}{3\cdot a}=\dfrac{c+a-b}{3\cdot b}\)
b)Tính giá trị biểu thức
\(P=\left(1+\dfrac{b}{a}\right)\cdot\left(1+\dfrac{a}{c}\right)\cdot\left(1+\dfrac{c}{b}\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}=\dfrac{a+b-c+b+c-a+c+a-b}{3a+3b+3c}=\dfrac{a+b+c+\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{3a+3b+3c}=\dfrac{a+b+c}{3\left(a+b+c\right)}=\dfrac{1}{3}\)
Khi đó:
\(\left\{{}\begin{matrix}\dfrac{a+b-c}{3c}=\dfrac{1}{3}\\\dfrac{b+c-a}{3a}=\dfrac{1}{3}\\\dfrac{c+a-b}{3b}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b-3c=3c\\3b+3c-3a=3a\\3c+3a-3b=3b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6c\\3b+3c=6a\\3c+3a=6b\end{matrix}\right.\)Thay vào \(P\)
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)
\(27P=3\left(\dfrac{a+b}{a}\right).3\left(\dfrac{c+a}{c}\right).3\left(\dfrac{b+c}{b}\right)\)
\(27P=\left(\dfrac{3a+3b}{a}\right)\left(\dfrac{3c+3a}{c}\right)\left(\dfrac{3b+3c}{b}\right)\)
\(27P=\)\(\dfrac{6c}{a}.\dfrac{6b}{c}.\dfrac{6a}{b}=\dfrac{216abc}{abc}=216\Leftrightarrow P=\dfrac{216}{27}=8\)
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Cho số 4,b,c khác 0 thỏa mãn \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
Tính P\(\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
<=> \(\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1\)
<=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
<=> a + b + c = 0 hoặc a = b = c.
Th1: a + b + c = 0
=> a + b = - c ; a + c = -b ; b + c = -a.
Thế vào P :
\(P=\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
\(=\left(\frac{a+b}{b}\right)\cdot\left(\frac{b+c}{c}\right)\cdot\left(\frac{c+a}{a}\right)\)
\(=-\frac{c}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)
TH2: a = b = c. THế vào P
\(P=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
Vậy: P = -1 nếu a + b + c = 0
hoặc P = 8 nếu a = b = c.
\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1=\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow P=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow a=b=c\)
\(\Rightarrow\hept{\begin{cases}a+b=2b\\b+c=2c\\c+a=2a\end{cases}}\)\(\Rightarrow P=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=2.2.2=8\)
Vậy \(P=-1\)hoặc \(P=8\)
rút gọn biểu thức sau bằng cách nhanh nhấtA left(a^2+b^2-c^2right)^2-left(a^2-b^2+c^2right)^2-4a^2b^2B left(3x^3+3x+1right)cdotleft(3x^3-3x+1right)-left(3x^3+1right)^2C left(2-6xright)^2+left(2-5xright)^2+2cdotleft(6x-2right)cdotleft(2-5xright)D 5cdotleft(3x-1right)^2+4cdotleft(5x+1right)^2-12cdotleft(5x-2right)left(5x+2right)E left(3x-1right)^2+left(2x+4right)cdotleft(1-3xright)+left(x+2right)^2G left(x-1right)^3+4cdotleft(x+1right)cdotleft(1-xright)+3cdotleft(x-1right)cdotleft(x^2+x+1rig...
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rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
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