\((\frac23x-3):(-10)=\frac25\)
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a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
\(a,\frac{-3}{2}-2x+\frac{3}{4}=-1\)
\(\frac{-3}{2}-2x=-1-\frac{3}{4}\)
\(\frac{-3}{2}-2x=\frac{-7}{4}\)
\(2x=\frac{-7}{4}+\frac{-3}{2}\)
\(2x=\frac{-13}{4}\)
\(x=\frac{-13}{4}:2\)
\(x=\frac{-13}{4}.\frac{1}{2}\)
\(x=\frac{-13}{8}\)
\(2\left(x+3\right)-x^2-3x=0\)
<=> \(2\left(x+3\right)-x\left(x+3\right)=0\)
<=> \(\left(x+3\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
\(2x\left(3x-5\right)=10-6x\)
<=> \(2x\left(3x-5\right)-\left(10-6x\right)=0\)
<=> \(2x\left(3x-5\right)-2\left(5-3x\right)=0\)
<=> \(2x\left(3x-5\right)+2\left(3x-5\right)=0\)
<=> \(2\left(3x-5\right)\left(x+1\right)=0\)
<=> \(\orbr{\begin{cases}3x-5=0\\x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\)\(2x+6-x^2-3x=0\)
\(\Leftrightarrow\)\(-x^2-x+6=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
\(2x\left(3x-5\right)=10-6x\)
\(\Leftrightarrow\)\(6x^2-10x=10-6x\)
\(\Leftrightarrow\)\(6x^2-4x-10=0\)
\(\Leftrightarrow\)\(2\left(x+1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\3x-5=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
Vậy....
a) \(x=\dfrac{25}{72}\)
b)\(x=-\dfrac{1}{4}\)
\(x=\dfrac{3}{2}\)
c)\(x=\dfrac{5}{4}\) hoặc
x \(=\dfrac{8}{5}\)
d và e chịu vì mk kg giỏi lắm về mũ
f)\(x=-2\)
G)\(x=-\dfrac{5}{12}\)
\(\left(\frac23x-3\right):\left(-10\right)=\frac25\)
\(\frac23x-3=\frac25\cdot\left(-10\right)\)
\(\frac23x-3=\frac{2\cdot\left(-10\right)}{5}\)
\(\frac23x-3=-\frac{20}{5}=-4\)
\(\frac23x=-4+3\)
\(\frac23x=1\)
\(x=1:\frac23\)
\(x=1\cdot\frac32\)
\(x=\frac32\)
Vậy \(x=\frac32\)
\(\left(\frac23x-3\right):\left(-10\right)=\frac25\)
\(\frac23x-3=-4\)
\(\frac23x=-1\)
\(x=-\frac32=-1,5\)
Vậy \(x=-1,5\)