Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?
Bạn kiểm tra lại nha
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
Từ a/b=c/d⇒a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau
a/c=b/d=a+b/c+d
⇒a^3/c^3=b^3/d^3=(a+b)^3/(c+d)^3 (1)
Từ a^3/c^3=b^3/d^3=a^3-b^3/c^3-d^3 (2)
Từ (1) và (2)
⇒(a+b)^3/(c+d)^3=a^3-b^3/c^3-d^3
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+............+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+..........+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+.........+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+.....+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+.........+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
=> \(7-3x=0\) hoặc \(2x+1=0\)
\(3x=7-0\) hoặc \(2x=0-1\)
\(3x=7\) hoặc \(2x=-1\)
\(x=7:3\) hoặc \(x=-1:2\)
\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)
Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)
Từ \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\\\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{1}{a}+\dfrac{1}{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Rightarrow a=b=c\)
Khi đó \(P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{3a^3}{3a^3}=1\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(A=1-\dfrac{1}{2^{100}}< 1\)
Vậy A < B.
Giải:
Có: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)
Lấy vế trừ theo vế, ta được:
\(A-\dfrac{1}{2}A=\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^{101}}}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^{100}}\right)}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
Lại có \(B=1\)
Vì \(1-\dfrac{1}{2^{100}}< 1\)
Nên \(A< B\)
Vậy \(A< B\).
Chúc bạn học tốt!
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).
B