a)Tìm số nguyên âm a,biết:
a^2-(3/5)^2=1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19+1/19.11+1/11.25
b)So sánh: C=1.3.5.7....99 với D=51/9.52/2.53/2....100/2
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T
0
MD
1
27 tháng 5 2021
Giải:
\(A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
Gọi: \(B=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
\(B=\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}+\dfrac{1}{19.22}+\dfrac{1}{22.25}\right):\dfrac{1}{2}\) \(B=\left[\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{19.22}+\dfrac{3}{22.25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\dfrac{24}{25}\right]:\dfrac{1}{2}\)
\(B=\dfrac{8}{25}:\dfrac{1}{2}\)
\(B=\dfrac{16}{25}\)
\(\Rightarrow A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(A^2=\dfrac{16}{25}+\dfrac{9}{25}\)
\(A^2=1\)
\(\Rightarrow A^2=1^2\) hoặc \(A^2=\left(-1\right)^2\)
\(A=1\) hoặc \(A=-1\)
Chúc bạn học tốt!
NT
0
VH
1
HT
3 tháng 9 2016
\(a^2-\frac{3}{5^2}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)
\(a^2-\frac{3}{5^2}=2.\left(\frac{1}{2.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}+\frac{1}{19.22}+\frac{1}{22.25}\right)\)
\(a^2-\frac{3}{5^2}=2.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{22}-\frac{1}{25}\right)\)
\(a^2-\frac{3}{5^2}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{25}\right)\)
=> \(a^2-\frac{3}{25}=\frac{2}{3}.\frac{23}{50}=\frac{23}{75}\)
=> \(a^2=\frac{23}{75}+\frac{3}{25}=\frac{32}{75}\)
=> \(a=\sqrt{\frac{32}{75}}\)(Nếu thế thì đây phải là đề của lớp 7 chứ nhỉ)
GH
2
4 tháng 9 2021
a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
AH
Akai Haruma
Giáo viên
4 tháng 9 2021
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
