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\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Rightarrow\left(a^{100}+b^{100}\right)\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)
\(\Rightarrow a^{202}+b^{202}+a^{100}b^{102}+a^{102}b^{100}=a^{202}+b^{202}+2a^{101}b^{101}\)
\(\Rightarrow a^{100}b^{100}\left(a^2+b^2\right)=a^{100}b^{100}\left(2ab\right)\)
\(\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a=b\)
Thế vào \(a^{100}+b^{100}=a^{101}+b^{101}\)
\(\Rightarrow a^{100}+a^{100}=a^{101}+a^{101}\)
\(\Rightarrow2a^{100}\left(a-1\right)=0\)
\(\Rightarrow a=1\Rightarrow b=1\)
\(\Rightarrow...\)
Ta có :
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{ab-bc}{\left(a+b\right)-\left(b+c\right)}=\frac{bc-ca}{\left(b+c\right)-\left(c+a\right)}=\frac{ab-ca}{\left(a+b\right)-\left(c+a\right)}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow Q=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=1\)
B1:
Ta có: a - b = ab => a = ab + b = b(a + 1)
Thay a = b(a + 1) vào a - b = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)
=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1
Lại có: ab = a - b
<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2
Vậy...
B2:
a, \(3y\left(y-\frac{2}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)
b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)
\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)
B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)
\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)
\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)
\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\), suy ra \(a=bk;c=dk\)
\(VT=\frac{2b^2k^2-3b^2k+3b^2}{2b^2+3b^2k}=\frac{b^2\left(2k^2-3k+3\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+3}{3k+2}\left(1\right)\)
\(VP=\frac{2d^2k^2-3d^2k+3d^2}{2d^2+3d^2k}=\frac{d^2\left(2k^2-3k+3\right)}{d^2\left(2+3k\right)}=\frac{2k^2-3k+3}{3k+2}\left(2\right)\)
Từ (1) và (2) suy ra ĐPcm
