Nếu \(a=b+c\) thì: \(\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}\)
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để chứng minh 1 trong 3 số a,b,c là lập phương của 1 số hữu tỉ ta sẽ chứng minh \(\sqrt[3]{a};\sqrt[3]{b};\sqrt[3]{c}\) có ít nhất 1 số hữu tỉ
đặt \(\hept{\begin{cases}x=\frac{a}{b^3}\\y=\frac{b}{c^3}\\z=\frac{c}{a^3}\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{x}=\frac{b^3}{a}\\\frac{1}{y}=\frac{c^3}{b}\\\frac{1}{z}=\frac{a^3}{b}\end{cases}}}\)
do abc=1 => xyz=1 (1)
từ đề bài => \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Rightarrow x+y+z=xy+yz+xz\left(xyz\ge1\right)\left(2\right)\)
Từ (1)(2) => \(xyz+\left(x+y+z\right)-\left(xy+yz+zx\right)-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
vậy \( {\displaystyle \displaystyle \sum }x=1 \) chẳng hạn, => \(a=b^3\)
\(\Rightarrow\sqrt[3]{a}=b\)mà b là số hữu tỉ
Vậy trong 3 số \(\sqrt[3]{a};\sqrt[3]{b};\sqrt[3]{c}\)có ít nhất 1 số hữu tỉ (đpcm)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\left(k\in Z\right)\)
\(\Rightarrow a=bk,c=dk\)
Có :
\(\left(\frac{a+b}{c+d}\right)^3=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{\left[b\left(k+1\right)\right]^3}{\left[d\left(k+1\right)\right]^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)
\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\left(=\frac{b^3}{d^3}\right)\)
Vậy ...
a/b = c/d =K ( K thuộc N* )
a = bK
c = dK
thay vào 2 cái cần so sanh đó là ok
k cho mik nha
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Ta có: \(\frac{a^3+b^3+c^3}{3}.\frac{a^2+b^2+c^2}{2}=\frac{a^5+b^5+c^5+a^3\left(b^2+c^2\right)+b^3\left(a^2+c^2\right)+c^3\left(a^2+b^2\right)}{6}\)
\(=\frac{a^5+b^5+c^5+a^3\left(\left(b+c\right)^2-2bc\right)+b^3\left(\left(c+a\right)^2-2ca\right)+c^3\left(\left(a+b\right)^2-2ab\right)}{6}\)
\(=\frac{a^5+b^5+c^5+a^3\left(a^2-2bc\right)+b^3\left(b^2-2ca\right)+c^3\left(c^2-2ab\right)}{6}\)
\(=\frac{\left(a^5+b^5+c^5\right)-abc\left(a^2+b^2+c^2\right)}{3}\)
Ma ta lại có:
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\frac{a^3+b^3+c^3}{3}.\frac{a^2+b^2+c^2}{2}=\frac{3\left(a^5+b^5+c^5\right)-\left(a^3+b^3+c^3\right)\left(a^2+b^2+c^2\right)}{9}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{3}.\frac{a^2+b^2+c^2}{2}=\frac{3\left(a^5+b^5+c^5\right)-\left(a^3+b^3+c^3\right)\left(a^2+b^2+c^2\right)}{9}\)
\(\Leftrightarrow\frac{5\left(a^3+b^3+c^3\right)\left(a^2+b^2+c^2\right)}{18}=\frac{\left(a^5+b^5+c^5\right)}{3}\)
\(\Leftrightarrow\frac{\left(a^3+b^3+c^3\right)\left(a^2+b^2+c^2\right)}{6}=\frac{\left(a^5+b^5+c^5\right)}{5}\) (ĐPCM)
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a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)
Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
a)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(đpcm\right)\)
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Chứng minh phải k bạn
\(\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)
Thay a=b+c ta có : \(\frac{\left(b+c+b\right)\left[\left(b+c\right)^2-ab+b^2\right]}{\left(b+c+c\right)\left[\left(b+c\right)^2-ab+b^2\right]}\)
\(\frac{\left(2b+c\right)\left(b^2+2bc+c^2-ab+b^2\right)}{\left(b+2c\right)\left(b^2+2bc+c^2-ab+b^2\right)}\)
Đặt b+c=a lại : \(\frac{2b+c}{b+2c}=\frac{a+b}{b+c}\)\(\Leftrightarrow\frac{\left(a+b\right)\left(2b^2+2bc+c^2-ab\right)}{\left(b+c\right)\left(2b^2+2bc+c^2-ab\right)}\)
\(=\frac{a+b}{b+c}\)
=> đpcm
Bạn ơi \(\frac{a+b}{a+c}mà\)chứ đâu phải \(\frac{a+b}{b+c}\)
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a + b + c = 3
a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ 3/2
Ta có
a/( 1 + b^2 ) = a - ab^2/( 1 + b^2 ) ≥ a - ab^2/2b = a - ab/2
Tương tự ta có
b/( 1 + c^2 ) ≥ b - bc/2
c/( 1 + a^2 ) ≥ c - ac/2
Cộng vào ta có
a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ a + b + c - ( ab + bc + ac )/2 = 3 - ( ab + bc + ac )/2
Xét ab + bc + ac
Ta có
a^2 + b^2 ≥ 2ab
b^2 + c^2 ≥ 2bc
c^2 + a^2 ≥ 2ac
=> a^2 + b^2 + c^2 ≥ ab + bc + ac
<=> a^2 + b^2 + c^2 + 2ac + 2bc + 2ab ≥ 3( ab + ac + bc )
<=> ( a + b + c )^2 ≥ 3( ab + ac + bc )
<=> ab + ac + bc ≤ 9:3 = 3
=> 3 - ( ab + bc + ac )/2 ≥ 3 - 3/2 = 3/2
=> a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥
Theo dãy tỉ số bằng nhau , có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}\)
\(\Rightarrow\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=2\)
\(\Rightarrow\left(\frac{b+c}{a}\right)^3=\left(\frac{c+a}{b}\right)^3=\left(\frac{a+b}{c}\right)^3=8\)
\(\Rightarrow\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}+\frac{\left(a+b\right)^3}{c^3}=8.3=24\)
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Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
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1,
\(\frac{a+2}{a-2}=\frac{b+3}{b-3}\)
<=> (a - 2)(b + 3) = (a + 2)(b - 3)
<=> ab + 3a - 2b - 6 = ab - 3a + 2b - 6
<=> 3a - 2b = -3a + 2b
<=> 6a = 4b
<=> 3a = 2b
<=> \(\frac{a}{2}=\frac{b}{3}\)(Đpcm)
2,
Có:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(=\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)
\(=\frac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}=0\)
=> bz - cy = 0
=> bz = cy
=> \(\frac{b}{y}=\frac{c}{z}\)(1)
=> cx - az = 0
=> cx = az
=> \(\frac{c}{z}=\frac{a}{x}\)(2)
Từ (1) và (2)
=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)(Đpcm)
thay vào như vậy: \(\frac{\left(b+c\right)^3+b^3}{\left(b+c\right)^3+c^3}=\frac{b+c+b}{b+c+c}\)
rồi sử dụng hằng đẳng thức lớp 8 ấy ạ