Cho a,b,c khác 0 thỏa mãn b2=\(a\times c\)
\(\dfrac{a}{bc}=\dfrac{\left(2010\times a+2011\times6\right)^2}{\left(2010\times b+2011\times c\right)^2}\)
Cho a,b,c>0 va abc=1 cmr
\(\dfrac{1}{a^3\times\left(b+c\right)}+\dfrac{1}{b^3\times\left(a+c\right)}+\dfrac{1}{c^3\times\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(=\dfrac{abc}{a^3\left(b+c\right)}+\dfrac{abc}{b^3\left(a+c\right)}+\dfrac{abc}{c^3\left(a+b\right)}\)
\(=\dfrac{bc}{a^2\left(b+c\right)}+\dfrac{ac}{b^2\left(a+c\right)}+\dfrac{ab}{c^2\left(a+b\right)}\)
\(=\dfrac{b^2c^2}{a^2bc\left(b+c\right)}+\dfrac{a^2c^2}{ab^2c\left(a+c\right)}+\dfrac{a^2b^2}{abc^2\left(a+b\right)}\)
\(Cauchy-Schwarz:\)
\(VT\ge\dfrac{\left(bc+ac+ab\right)^2}{abc\left[a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)\right]}\)
\(=\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\)
\(AM-GM:\)
\(ab+bc+ca\ge\sqrt[3]{\left(abc\right)^2}=3\)
\(\Rightarrow VT\ge\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
Lời giải khác:
Áp dụng BĐT AM-GM:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)
\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq 2\sqrt{\frac{1}{4b^2}}=\frac{1}{b}=\frac{abc}{b}=ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq 2\sqrt{\frac{1}{4c^2}}=\frac{1}{c}=\frac{abc}{c}=ab\)
Cộng theo vế và rút gọn:
\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}+\frac{ab+bc+ac}{2}\ge ab+bc+ac\)
\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\) (AM_GM)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
a) Cho các số a,b,c,d khác 0 . Tính :
T = \(x^{2011}+y^{2011}+z^{2011}+t^{2011}\)
Biết x,y,z,t thoả mãn \(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}\)
b) Tìm số tự nhiên M nhỏ nhất có 4 chữ số thoả mãn điều kiện
M=a+b=c+d=e+f
Nếu câu b thiếu j thì các bạn cứ bỏ qua nha
Cho a, b, c thỏa mãn \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\)
Tính \(M=\left(a-b\right)\left(b-c\right)-c^2-2ac-a^2\)
>; <; =?
a) \(\dfrac{2}{3}\times\dfrac{4}{5}\) \(\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}+\dfrac{2}{15}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
A = \(\dfrac{-19}{9}\times\dfrac{1}{2}-\dfrac{4}{11}\times\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)\)
B = \(\left(-\dfrac{15}{6}\right)\div\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}\times\dfrac{-11}{2}\)
C = \(\dfrac{3}{4}\times\left(-8\right)-\dfrac{1}{3}\times\dfrac{-7}{2}-\dfrac{5}{18}\)
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
*) a,Cho các số a,b,c,d khác 0. Tính
T=\(x^{2011}+y^{2011}+z^{2011}+t^{2011}\)
Biết x,y,z,t thỏa mãn: \(\dfrac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}\)=\(\dfrac{x^{2010}}{a^2}+\dfrac{y^{2010}}{b^2}+\dfrac{z^{2010}}{c^2}+\dfrac{t^{2010}}{d^2}\)
b,Tìm sốtự nhiên M nhỏ nhất có 4 chữ số sao cho:
M = a+b=c+d=e+f
Biết a,b,c,d,e,f \(\in\) N* và \(\dfrac{a}{b}=\dfrac{14}{22};\dfrac{c}{d}=\dfrac{11}{13};\dfrac{e}{f}=\dfrac{17}{13}\)
c, Cho 3 số a,b,c thỏa mãn:\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\)
Tính giá trị của biểu thức M = 4(a - b)(b - c) - (c - a)\(^2\)
Câu c,
Đặt a/2009=b/2010=c/2011=k
=>a=2009.k (1)
b=2010.k (2)
c=2011.k (3)
Thay (1),(2),(3) vào biểu thức:
M=4.(a-b)(b-c)-(c-a)^2 ,ta được:
M=4.(2009.k-2010.k)(2010.k-2011.k)-(2011.k-2009.k)^2
M=4.(-k).(-k)-(2k)^2
M=4.k^2-4.k^2
M=0.
a,\(\dfrac{1}{3}\times\left(x-1\right)+\dfrac{2}{5}\times\left(x+1\right)=0\)
b,4x-\(\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}x-5\right)\)
c,\(\left(x+\dfrac{1}{2}\right)\times\left(x-\dfrac{3}{4}\right)=0\)
Các bn ơi giúp mk với chiều mk đi học rồi !!!!!!!!!!
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
Tìm x
a, \(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
b, \(\left(x-\dfrac{3}{2}\right)\times\left(2\times x-1\right)>0\)
c, \(\left(x-\dfrac{4}{7}\right)\div\left(x+\dfrac{1}{2}\right)>0\)
d, \(\left(2\times x-5\right)\div\left(x+1\dfrac{3}{4}\right)< 0\)
Giúp mk với
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
cho abs số a,b,c thỏa mãn \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\)tính giá trị của biểu thức \(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
Ta có \(\dfrac{a}{2009}\)=\(\dfrac{b}{2010}\)=\(\dfrac{c}{2011}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{c-a}{2011-2009}=\dfrac{c-a}{2}\left(1\right)\)
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{a-b}{2009-2010}=\dfrac{a-b}{-1}\)(2)\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{b-c}{2010-2011}=\dfrac{b-c}{-1}\left(3\right)\)
Từ (1),(2),(3) \(_{\Rightarrow}\)\(\dfrac{c-a}{2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\Rightarrow\dfrac{\left(a-c\right)^{ }2}{2^{ }2}=\dfrac{\left(a-b\right)}{-1}\times\dfrac{\left(b-c\right)}{-1}\)
\(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(a-b\right)\times\left(b-c\right)}{1}\Rightarrow4\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\)
\(\Rightarrow M=4\left(a-b\right).\left(a-c\right)-\left(c-a\right)^2=0\)
Vậy M = 0
đặt \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=k\) ta có:
\(\Rightarrow a=2009k\left(1\right)\\ \Rightarrow b=2010k\left(2\right)\\ \Rightarrow c=2011k\left(3\right)\)
thay 1;2;3 vào M ta có:
\(M=4\left(2009k-2010k\right)\left(2010k-2011k\right)-\left(2011k-2009k\right)^2\\ \Rightarrow M=4.\left(-k\right)\left(-k\right)-\left(2k\right)^2\\ \Rightarrow M=4k^2-\left(2k\right)^2\\ \Rightarrow M=\left(2k\right)^2-\left(2k\right)^2\\ \Rightarrow M=0\)Vậy M = 0