a)Tính 1-1/2;1/2-1/3;1/3-1/4;1/4-1/5;1/5-1/6
b) Sử dụng kết quả trên để tính nhanh tổng :
S=1/2+1/6+1/12+1/20+1/30
Nhớ ghi cách giải nữa nhé
Ai nhanh mình tích cho
Giúp mình với
K
Khách
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Những câu hỏi liên quan
NH
1
23 tháng 12 2020
uses crt;
var a,m,i:integer;
s:real;
begin
clrscr;
write('Nhap a='); readln(a);
write('Nhap m='); readln(m);
s:=1;
for i:=1 to m do
s:=s+1/sqr(a+i);
writeln(s:4:2);
readln;
end.
AH
Akai Haruma
Giáo viên
28 tháng 8 2021
Lời giải:
a. ĐKXĐ: $a\geq 0; a\neq 1$
b.
\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}+1\right].\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-1\right].\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\)
\(=(\sqrt{a}+1)(\sqrt{a}-1).\sqrt{2}=\sqrt{2}(a-1)\)
c.
\(P=\sqrt{2}(\sqrt{2+\sqrt{2}}-1)=\sqrt{4+2\sqrt{2}}-\sqrt{2}\)
NT
28 tháng 8 2021
a. ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne1\\\sqrt{a}\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
b. \(P=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right).\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right].\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right].\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right).\sqrt{2}=2\left(a-1\right)=2a-2\)
16 tháng 11 2021
\(a,P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\left(-1< a< 1\right)\\ P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{2+\sqrt{\left(1-a\right)\left(1+a\right)}}\\ P=\sqrt{1-a}\\ b,a=\dfrac{24}{49}\Leftrightarrow1-a=\dfrac{25}{49}\\ \Leftrightarrow P=\sqrt{1-a}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\\ c,P=2\Leftrightarrow1-a=4\Leftrightarrow a=-3\left(ktm\right)\Leftrightarrow a\in\varnothing\)
C
1
10 tháng 1 2022
\(a,A=\dfrac{x^2-x-2}{x^2-1}+\dfrac{1}{x-1}-\dfrac{1}{x+1}\)
\(\Rightarrow A=\dfrac{x^2-x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-x-2x+x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-2x-x+2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x-2}{x+1}\)
\(b,A=\dfrac{3}{4}\\ \Rightarrow\dfrac{x-2}{x+1}=\dfrac{3}{4}\\ \Rightarrow4\left(x-2\right)=3\left(x+1\right)\\ \Rightarrow4x-8=3x+3\\ \Rightarrow4x-8-3x-3=0\\ \Rightarrow x-11=0\\ \Rightarrow x=11\)
\(c,\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Thay x=5 vào A ta có:
\(A=\dfrac{x-2}{x+1}=\dfrac{5-2}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)
Thay x=1 vào A ta có:
\(A=\dfrac{x-2}{x+1}=\dfrac{1-2}{1+1}=\dfrac{-1}{2}\)