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b) Giải:
ĐK: \(a\ne-b\)
Ta có:
\(3a^2+b^2=4ab\)
\(\Leftrightarrow4a^2-4ab+b^2-a^2=0\)
\(\Leftrightarrow\left(2a-b\right)^2-a^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\\a=b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{b}{3}\Leftrightarrow P=\dfrac{\dfrac{b}{3}-b}{\dfrac{b}{3}+b}=\dfrac{-1}{2}\\a=b\Leftrightarrow P=\dfrac{a-a}{a+a}=\dfrac{0}{2a}=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}P=\dfrac{-1}{2}\\P=0\end{matrix}\right.\)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\)
\(2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\)
\(2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)\)
\(B=1-\dfrac{1}{2^{99}}\)
\(B< 1\)
\(\Rightarrowđpcm\)
bài 3 : \(\left\{{}\begin{matrix}ab=2\\bc=3\\ca=54\end{matrix}\right.\)
hiển nhiên a;b;c =0 không phải nghiệm
\(\Leftrightarrow\left(abc\right)^2=2.3.54=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}abc=-18\\abc=18\end{matrix}\right.\)
abc=-18 => c=-9; a=-6; b=-1/3
abc=18 => c=9; a=6; b=1/3
Lời giải:
\(A=\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+...+(\frac{1}{2})^{98}+(\frac{1}{2})^{99}\)
\(\Rightarrow 2A=1+\frac{1}{2}+(\frac{1}{2})^2+...+(\frac{1}{2})^{97}+(\frac{1}{2})^{98}\)
Trừ theo vế:
\(2A-A=1-(\frac{1}{2})^{99}\)
\(A=1-(\frac{1}{2})^{99}< 1\)
Ta có đpcm.
ta có : \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow\dfrac{1}{2}B=\dfrac{1}{2}\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\right)=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow B-\dfrac{1}{2}B=\dfrac{1}{2}B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow B=2.\dfrac{1}{2}B=1\left(\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\right)=1-\left(\dfrac{1}{2}\right)^{99}< 1\)
vậy \(B< 1\)
2B= 1+ 1/2+ (1/2)2+ ....+(1/2)98
_
B= 1/2+ (1/2)2+ ....+(1/2)99
B= 1- (1/2)99 <1
=>B <1
bai 1
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)
\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
Lời giải:
Ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow \frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{100}\)
Trừ theo vế:
\(\Rightarrow \frac{B}{2}=\left(\frac{1}{2}\right)^{100}-\frac{1}{2}\)
\(\Leftrightarrow B=\left(\frac{1}{2}\right)^{99}-1<2-1\Leftrightarrow B< 1\)
Vì \(\left(\frac{1}{2}\right)^{99}\not\in\mathbb{Z};1\in\mathbb{Z}\Rightarrow B\not\in \mathbb{Z}\)
Ta có đpcm.
\(PHUCDZ=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(PHUCDZ=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2PHUCDZ=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)
\(2PHUCDZ=1+\dfrac{1}{2}+...+\dfrac{1}{2^{98}}\)
\(2PHUCDZ-PHUCDZ=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)
\(PHUCDZ=1-\dfrac{1}{2^{99}}< 1\)
\(\Rightarrowđpcm\)
\(PHUCDZ=1-\dfrac{1}{2^{99}}=\dfrac{2^{99}}{2^{99}}-\dfrac{1}{2^{99}}=\dfrac{2^{99}-1}{2^{99}}\)
Vì \(2^{99}-1\) và \(2^{99}\) là 2 số nguyên tố cùng nhau nên không thể rút gọn cho 1 số nào khác 1.
Vậy \(PHUCDZ\ne Z\Rightarrowđpcm\)