tìm x
a)(x-3)^2+x(10-x)=7
b)(x-2)^2-x(x-1)=13
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HN
6
DC
1
17 tháng 10 2021
a: Ta có: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow x^2+5x-10=x^2+3x-4\)
\(\Leftrightarrow2x=6\)
hay x=3
7 tháng 7 2023
a: =16-2+91=14+91=105
b: =9*5+8*10-27=45+53=98
c: =32+65-3*8=8+65=73
d; \(=5^3-10^2=125-100=25\)
e: \(=4^2-3^2+1=8\)
f: =9*16-16*8-8+16*4
=16(9-8+4)-8
=16*5-8
=72
H9
HT.Phong (9A5)
CTVHS
7 tháng 7 2023
a) \(2^4-50:25+13\cdot7\)
\(=2^4-2+91\)
\(=16-2+91\)
\(=14+91\)
\(=105\)
b) \(3^2\cdot5+2^3\cdot10-3^4:3\)
\(=9\cdot5+8\cdot10-3^3\)
\(=45+80-27\)
\(=98\)
c) \(2^5+5\cdot13-3\cdot2^3\)
\(=32+65-3\cdot8\)
\(=32+65-24\)
\(=73\)
d) \(5^{13}:5^{10}-5^2\cdot2^2\)
\(=5^{13-10}-\left(5\cdot2\right)^2\)
\(=5^3-10^2\)
\(=125-100\)
\(=25\)
e) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^{5-3}-3^{9-7}+1\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=8\)
f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)
\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)
\(=4^2\cdot\left(3^2-2^3+2^2\right)-2^3\)
\(=4^2\cdot\left(9-8+4\right)-8\)
\(=16\cdot5-8\)
\(=72\)
Y
27 tháng 1 2023
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
28 tháng 12 2021
Bài 1:
a) \(2\)\(\dfrac{2}{3}\)\(=\dfrac{8}{3}\)
b) \(1\)\(\dfrac{5}{7}=\dfrac{12}{7}\)
Đã lên đại úy gòi
MH
0
MH
0
Tìm x:
a)(x-3)2+x(10-x)=7
x2-6x+9+10x-x2=7
4x+9=7
4x=-2
=>x=-0,5
b)(x-2)2-x(x-1)=13
x2-4x+4-x2+x=13
-3x+4=13
-3x=9
=>x=-3