tính hợp lí
8x9x14+6x7x12+19x4x18
98 x 31 + 62
1 + 3 + 5 + 7+ .....+ 27 + 29
tim x
X x ( x - 10 ) = 0
( 2 x X - 4 ) x ( X + 2 ) = 0
mk cần gấp giúp mk nha thanks những bạn nào giải hộ mk mk sẽ tích cho
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LN
27 tháng 2 2018
Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{16}=\frac{2x}{16}=\frac{x}{8}=\frac{25x}{200}=\frac{xy}{200}\)
Suy ra: \(25x=xy\Rightarrow y=25\)
Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}\)
Suy ra: \(13x-13y=3x+3y\)
Thế y vào đẳng thức trên:
\(13x-325=3x+75\)
Suy ra: \(10x=325+75=400\Rightarrow x=40\)
Vậy ........
HN
3 tháng 7 2018
a) \(\left(x-3\right)^2.\left(x+3\right).\left(x-3\right)\)
\(=\left(x-3\right).\left(x-3\right).\left(x+3\right).\left(x-3\right)\)
\(=\left(x-3\right)^3.\left(x+3\right)\)
\(=\left(3x-9\right).\left(x+3\right)\)
Phần b tương tự
2 tháng 1 2019
X-3/5=1+2/3 4/7:X=1/2x2/5
X-3/5=5/3 4/7:X=1/5
X=5/3+3/5 X=4/7:1/5
X=34/15 X=4
Vậy X=34/15 Vậy X=4
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
4 tháng 2 2017
\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)
\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)
\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)
\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)
\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)
\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)
\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\) ( 4^3 = 8^2 ; 5^4 = 25^2 )
\(=\frac{1}{25\times7}=\frac{1}{175}\)
4 tháng 2 2017
\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)
\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)
\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)
\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)
Q
4 tháng 6 2017
a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Leftrightarrow9x-14x-8=5\)
\(\Leftrightarrow-5x-8=5\)
\(\Leftrightarrow-5x=5+8\)
\(\Leftrightarrow-5x=13\)
\(\Rightarrow x=-\dfrac{13}{5}\)
Vậy \(x=-\dfrac{13}{5}\)
b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)
\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)
đến đây bạn giải tiếp nhé
c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)
