A=\(\frac{x}{7-x}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
26 tháng 11 2016
\(a,97\times327+327\times3\)
\(=327\times\left(97+3\right)\)
\(=327\times100\)
\(=32700\)
\(b,\frac{1}{7}\times\frac{4}{23}+\frac{1}{7}\times\frac{25}{23}+\frac{1}{7}\times\frac{17}{23}+5\times\frac{1}{7}\)
\(=\frac{1}{7}\times\left(\frac{4}{23}+\frac{25}{23}+\frac{17}{23}+5\right)\)
\(=\frac{1}{7}\times7\)
\(=1\)
S
26 tháng 11 2016
a) 97 x 327 + 327 x 3
= (97 + 3) x 327
= 100 x 327 = 32700
b) \(\frac{1}{7}\times\frac{4}{23}+\frac{1}{7}\times\frac{25}{23}+\frac{1}{7}\times\frac{17}{23}+5\times\frac{1}{7}\)
\(=\frac{1}{7}\times\left(\frac{4}{23}+\frac{25}{23}+\frac{17}{23}+5\right)\)
\(=\frac{1}{7}\times7=\frac{7}{7}=1\)
28 tháng 8 2017
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
LN
12 tháng 10 2018
Ta có:
\(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
\(\Leftrightarrow\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2010}+\frac{x+2017}{2011}=\frac{x+2017}{2012}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}\right)\)
Suy ra \(x+2017=0\)
Vậy \(x=-2017\)
b) Dễ tự làm nhé
31 tháng 12 2016
Bài 1:
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}=\frac{5}{7}\)
Bài 2:
a) \(\frac{x}{7}+\left(\frac{-3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(\Rightarrow\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)
b) \(\left(x-1\right)^{x+6}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+4}=0\)
\(\Rightarrow\left(x-1\right)^{x+4}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(x-1\right)^{x+1}=0\) hoặc \(\left(x-1\right)^2-1=0\)
+) \(\left(x-1\right)^{x+1}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(x-1\right)^2-1=0\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\left(x-1\right)=\pm1\)
+ \(x-1=1\Rightarrow x=2\)
+ \(x-1=-1\Rightarrow x=0\)
Vậy \(x\in\left\{0;2;1\right\}\)
HN
31 tháng 12 2016
1)
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)
\(=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}\)
\(=\frac{5}{7}\)
2) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)x - \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right) = \dfrac{9}{{20}}\\x = \dfrac{9}{{20}} + \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right)\\x = \dfrac{9}{{20}} + \dfrac{{25}}{{20}} - \dfrac{{28}}{{20}}\\x = \dfrac{{6}}{{20}}\\x = \dfrac{{ 3}}{{10}}\end{array}\)
Vậy \(x = \dfrac{{ 3}}{{10}}\)
\(\begin{array}{*{20}{l}}{b)9 - x = \dfrac{8}{7} - \left( { - \dfrac{7}{8}} \right)}\\\begin{array}{l}9 - x = \dfrac{8}{7} + \dfrac{7}{8}\\9 - x = \dfrac{{64}}{{56}} + \dfrac{{49}}{{56}}\\9 - x = \dfrac{{113}}{{56}}\end{array}\\{x = 9 - \dfrac{{113}}{{56}}}\\{x = \dfrac{{504}}{{56}} - \dfrac{{113}}{{56}}}\\{x = \dfrac{{391}}{{56}}}\end{array}\)
Vậy \(x = \dfrac{{391}}{{56}}\)
16 tháng 12 2016
a) ĐK: \(x\ne\left\{0;\pm7;49\right\}\)
b) \(\left(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\right):\frac{2x-7}{x^2+7x}-\frac{x}{x-7}\)
Xét \(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\)
= \(\frac{x^2}{x\left(x-7\right)\left(x+7\right)}-\frac{\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x^2-49\right)}{x.\left(x-7\right)\left(x+7\right)}\)
=\(\frac{49}{x\left(x+7\right)\left(x-7\right)}\) (ĐK: x\(\ne\) { 0; 7;-7;49}
16 tháng 12 2016
sau đó: chia trước trừ sau
Đoạn sau dễ chắc bạn tự làm được
Làm bài tốt
D
2
10 tháng 10 2016
\(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)
\(\Rightarrow7x-7=6x+30\)
\(\Rightarrow7x-6x=30+7\)
\(\Rightarrow x=37\)
10 tháng 10 2016
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Rightarrow x^2+7x-2x+14=x^2-x+4x-4\)
\(\Rightarrow x^2-x^2+7x-4x+x-2x=-14-4\)
\(\Rightarrow2x=-18\)
\(\Rightarrow x=-\frac{18}{2}=-9\)
TN
23 tháng 3 2020
\(a.A=\left(\frac{x+2}{x^2-2x}+\frac{x-6}{x^2-4}-\frac{x-2}{x^2+2x}\right):\frac{2x-7}{x^2-4x+4}\\ \Leftrightarrow A=\left(\frac{x+2}{x\left(x-2\right)}+\frac{x-6}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{x\left(x+2\right)}\right):\frac{2x-7}{x^2-4x+4}\\ \Leftrightarrow A=\left(\frac{\left(x+2\right)\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}+\frac{x\left(x-6\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\right):\frac{2x-7}{x^2-4x+4}\\ \Leftrightarrow A=\left(\frac{\left(x+2\right)\left(x+2\right)+x\left(x-6\right)-\left(x-2\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\right):\frac{2x-7}{x^2-4x+4}\)
\(\Leftrightarrow A=\left(\frac{x^2+2x}{x\left(x+2\right)\left(x-2\right)}\right):\frac{2x-7}{x^2-4x+4}\\ \Leftrightarrow A=\frac{x\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}:\frac{2x-7}{x^2-4x+4}\\ \Leftrightarrow A=\frac{1}{x-2}:\frac{2x-7}{\left(x-2\right)^2}\\ \Leftrightarrow A=\frac{1}{x-2}.\frac{\left(x-2\right)^2}{2x-7}\\ \Leftrightarrow A=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(2x-7\right)}\\ \Leftrightarrow A=\frac{x-2}{2x-7}\)
đề là j?tính A ak?hay cmr A là ps tối giản............????????????
???? câu hỏi đâu