\(\frac{1}{a}-\frac{1}{b}=\frac{1}{a}.\frac{1}{b}\)
Tim a,b\(\inℚ\)
Cho a,b,c \(\inℚ\)
CMR \(\sqrt{\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\inℚ}\)
\(Cho\)\(a,b,c\ne0,\inℚ\)và \(a=b+c\)
\(CMR:\)\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\inℚ\)
Ta có: \(\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2+2\left(\frac{1}{ab}+\frac{1}{ac}-\frac{1}{bc}\right).\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{2}{ab}-\frac{2}{ac}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}-\frac{2}{bc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)(1)
Mặt khác \(\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2+2\left(\frac{1}{ab}+\frac{1}{ac}-\frac{1}{bc}\right)\)
\(=\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2+2.\frac{c+b-a}{abc}\)
\(=\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2\)(vì a=b+c) (2)
Từ (1) và (2) Suy ra
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2\)
\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}|.\)
Do a,b,c là các số hữu tỉ khác 0 nên \(|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}|\)là một số hữu tỉ
Từ đây ta có điều phải chứng minh
a) \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
b)\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)
Tim a,b biet\(a+b=3\left(a-b\right)=2\frac{a}{b}\)
\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
\(A=\frac{2^9-1}{2^9}\)
Vậy \(A=\frac{2^9-1}{2^9}\)
Chúc bạn học tốt ~
\(\frac{9}{11}=\frac{1}{a+\frac{1}{b+\frac{1}{c}}}\)
tim a,b,c
Ta có:\(\frac{9}{11}\)=\(\frac{1}{\frac{11}{9}}\)=\(\frac{1}{1+\frac{2}{9}}\)=\(\frac{1}{1+\frac{1}{\frac{9}{2}}}\)=\(\frac{1}{1+\frac{1}{4+\frac{1}{2}}}\).Từ đó suy ra:a=1;b=4;c=2
Tim GTNN của \(C=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\)với \(a\ge0,b\ge0,c\ge0\)và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le0\)
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
tim a;b;c;d
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
tim a, b, c, d
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)thế này à
tim a,b
\(a,\frac{1}{a}+\frac{2}{b}=\frac{1}{3}\)
\(b,\frac{1}{a}+\frac{3}{b}=\frac{3}{4}\)
Tìm \(y\inℚ\), biết :
a)\(|2y-3|-\frac{1}{7}=\frac{3}{4}\)
Tìm giá trị lớn nhất:
a)\(A=-|2x-5|+32\)
b)\(B=\frac{-2}{7}|y-\frac{1}{3}|\)
Tìm giá trị nhỏ nhất :
\(C=|y^2+1|+2020\)
a) \(\left|2y-3\right|-\frac{1}{7}=\frac{3}{4}\)
=> \(\left|2y-3\right|=\frac{3}{4}+\frac{1}{7}\)
=> \(\left|2y-3\right|=\frac{25}{28}\)
=> \(\orbr{\begin{cases}2y-3=\frac{25}{28}\\2y-3=-\frac{25}{28}\end{cases}}\)
=> \(\orbr{\begin{cases}2y=\frac{109}{28}\\2y=\frac{59}{28}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{109}{56}\\x=\frac{59}{56}\end{cases}}\)
Tính GTLN
a) Ta có: -|2x - 5| \(\le\)0 \(\forall\)x
=> -|2x - 5| + 32 \(\le\)32 \(\forall\)x
Hay A \(\le\)32 \(\forall\)x
Dấu "=" xảy ra khi : 2x - 5 = 0 <=> 2x = 5 <=> x = 5/2
Vậy Max của A = 32 tại x = 5/2
\(C=\left|y^2+1\right|+2020\)
Ta có: \(y^2\ge0\Leftrightarrow y^2+1\ge1\Leftrightarrow\left|y^2+1\right|\ge1\)
\(\Leftrightarrow C=\left|y^2+1\right|+2020\ge2021\)
Vậy \(C_{min}=2021\)
(Dấu "="\(\Leftrightarrow y^2+1=1\Leftrightarrow y^2=0\Leftrightarrow y=0\))
b) \(B=\frac{-2}{7}\left|y-\frac{1}{3}\right|\)
Ta có: \(\left|y-\frac{1}{3}\right|\ge0\)
\(\Leftrightarrow-\left|y-\frac{1}{3}\right|\le0\)
\(\Leftrightarrow\frac{-2}{7}\left|y-\frac{1}{3}\right|\le0\)
Vậy \(B_{min}=0\)
(Dấu "="\(\Leftrightarrow y-\frac{1}{3}=0\Leftrightarrow y=\frac{1}{3}\))