PH
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
YY
24 tháng 4 2018
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
PY
24 tháng 4 2018
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
2 tháng 4 2018
\(a)\) \(-\left(x+84\right)+213=-16\)
\(\Leftrightarrow\)\(-x-84+213=-16\)
\(\Leftrightarrow\)\(x=213-84+16\)
\(\Leftrightarrow\)\(x=145\)
Vậy \(x=145\)
\(b)\) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=2\)
Chúc bạn học tốt ~
NT
2 tháng 4 2018
a) \(-\left(x+84\right)+213=-16\)
\(-\left(x+84\right)=-16-213\)
\(-\left(x+84\right)=-229\)
\(\Rightarrow x+84=229\)
\(\Rightarrow x=229-84=145\)
Vậy \(x=145\)
b) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|\frac{-1}{4}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|-1\right|\)
\(\left(x-1\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1+1=2\\x=-1+1=0\end{cases}}\)
Vậy \(x\in\left\{0;2\right\}\)
2 tháng 8 2016
a) \(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^2.2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{9.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{9.2^{36}}{2^{35}.9}=2\)
b) \(\frac{3}{2}.x-\left(\frac{4}{5}-2.x\right)=1\frac{3}{10}:\frac{3}{2}\)
\(\frac{3}{2}.x-\frac{4}{5}+2.x=\frac{13}{10}:\frac{3}{2}\)
\(\left(\frac{3}{2}.x+2.x\right)-\frac{4}{5}=\frac{13}{10}.\frac{2}{3}\)
\(x.\left(\frac{3}{2}+2\right)-\frac{4}{5}=\frac{13}{15}\)
\(x.\frac{7}{2}=\frac{13}{15}+\frac{4}{5}\)
\(x.\frac{5}{2}=\frac{13}{15}+\frac{12}{15}\)
\(x.\frac{7}{2}=\frac{25}{15}=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{7}{2}\)
\(x=\frac{5}{3}.\frac{2}{7}=\frac{10}{21}\)
BT
22 tháng 3 2017
<=> (1-1/10)(x-1)+x/10=x-9/10
<=> 9x/10-9/10+x/10=x-9/10
<=> x=x
Như vậy, phương trình thỏa mãn với mọi x
S
18 tháng 5 2017
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
5 tháng 4 2017
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
a) (x - 1/2) x 2 = 9/16
=> x - 1/2 = 9/16 : 2
=> x - 1/2 = 9/16 x 1/2
=> x - 1/2 =9/32
=> x = 9/32 + 1/2
=> x = 25/32
b) |x + 1/2| = 3/4
=> x + 1/2 = 3/4 hoặc x + 1/2 =-3/4
=>x = 3/4 - 1/2 hoặc x = -3/4 -1/2
=>x = 1/4 hoặc x = -5/4
Vậy .........