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\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
Bài 3:
a) (3x - 2)(4x + 5)=0
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
vậy phương trình có tập ngiệm S={\(\dfrac{-5}{4};\dfrac{2}{3}\)}
b) 2x(x-3)-5(x-3)=0
\(\Leftrightarrow\left(x-3\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy phương trình có tập ngiệm S={\(3;\dfrac{5}{2}\)}
c) 2x(x + 3) + 5(x + 3)=0
\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{-5}{2}\end{matrix}\right.\)
vậy phương trình có tập ngiệm S={\(-3;\dfrac{-5}{2}\)}
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)
bài 2:
ĐKXĐ: x khác -1
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow0x=-1\)
\(\Leftrightarrow x\in\varnothing\)
Suy ra pt vô nghiệm
b.
ĐKXĐ: x khác \(\dfrac{3}{2}\)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
1)
a)
\(2x+5=20+3x\\ \Leftrightarrow2x+5-20-3x=0\\ \Leftrightarrow-x-15=0\\ \Rightarrow x=-15\)
b)
\(2.5y+1.5=2.7y-1.5c\cdot2t-\frac{3}{5}=\frac{2}{3}-t\\ \Leftrightarrow2.5y+1.5-2.7y+3ct+\frac{3}{5}-\frac{2}{3}+t=0\\ \Leftrightarrow-0.2y+\frac{43}{30}+3ct+t=0\)
2)
a)
\(\frac{5x-4}{2}=\frac{16x+1}{7}\\ \Leftrightarrow\frac{35x-28}{14}-\frac{32x+2}{14}=0\\ \Leftrightarrow\frac{35x-28-32x-2}{14}=0\\ \Leftrightarrow\frac{3x-30}{14}=0\\ \Rightarrow3x-30=0\\ \Rightarrow x=10\)
b)
\(\frac{12x+5}{3}=\frac{2x-7}{4}\\ \Leftrightarrow\frac{48x+20}{12}-\frac{6x-21}{14}=0\\ \Leftrightarrow\frac{48x+20-6x+21}{12}=0\\ \Leftrightarrow\frac{42x+41}{12}=0\\ \Rightarrow42x+41=0\\ \Rightarrow x=-\frac{41}{42}\)
3)
a)
\(\left(x-1\right)^2-9=0\\ \Leftrightarrow\left(x-1-3\right)\cdot\left(x-1+3\right)=0\\ \Leftrightarrow\left(x-4\right)\cdot\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Bài 2:
a, \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(2x-1\right)-5\left(x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3\right)-\left(5x+40\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-43=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{1;43\right\}\)
b, \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow9x^2-1-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{3};-2\right\}\)
c, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(49-x^2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(7-x\right)\left(7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1-7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\4x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-7;2\right\}\)
d, \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;2;3\right\}\)
e, \(2x^3+3x^2-32x=48\)
\(\Leftrightarrow2x^3+3x^2-32x-48=0\)
\(\Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\)
\(\Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{4;-4;3-\frac{3}{2}\right\}\)
a) \(3\left(2x-x\right)=5x+1\)
\(\Leftrightarrow6x-3x=5x+1\)
\(\Leftrightarrow6x-3x-5x=1\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\dfrac{1}{-2}=-\dfrac{1}{2}\)
b) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)
\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)
\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)
\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)
\(\Leftrightarrow x+2022=0\)
\(\Leftrightarrow x=-2022\)
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