tìm x
2(x+3)-x^2-3x=0
2x(3x-5)=10-6x
x^3-x^2=4x^2-8x+14
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12 tháng 10 2021
\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)
N
12 tháng 10 2021
\(\left(4-8x\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)
\(2\left(x-4\right)-3x\left(x-4\right)=0\)
\(\left(2-3x\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
2 tháng 10 2021
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
PN
17 tháng 5 2021
a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)
\(< =>8x^2+12x-8x^2+x=5x+10\)
\(< =>13x=5x+10< =>8x=10\)
\(< =>x=\frac{10}{8}=\frac{5}{4}\)
b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)
\(< =>9x^2-25-9x^2+x=4\)
\(< =>x=4+29=33\)
PN
17 tháng 5 2021
c,\(3-4x\left(25-2x\right)=8x^2+x-300\)
\(< =>3-100x+8x^2=8x^2+x-300\)
\(< =>x+100x=3+300\)
\(< =>101x=303< =>x=\frac{303}{101}=3\)
d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)
\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)
\(< =>\frac{30x-6x-24x+15-4}{20}=5\)
\(< =>\frac{11}{5}=5< =>11=25\)(vo li)
\(2\left(x+3\right)-x^2-3x=0\)
<=> \(2\left(x+3\right)-x\left(x+3\right)=0\)
<=> \(\left(x+3\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
\(2x\left(3x-5\right)=10-6x\)
<=> \(2x\left(3x-5\right)-\left(10-6x\right)=0\)
<=> \(2x\left(3x-5\right)-2\left(5-3x\right)=0\)
<=> \(2x\left(3x-5\right)+2\left(3x-5\right)=0\)
<=> \(2\left(3x-5\right)\left(x+1\right)=0\)
<=> \(\orbr{\begin{cases}3x-5=0\\x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\)\(2x+6-x^2-3x=0\)
\(\Leftrightarrow\)\(-x^2-x+6=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
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....