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a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
Trả lời:
\(1,3x\left(x-7\right)+2x-14=0\)
\(\Leftrightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{2}{3}\end{cases}}}\)
Vậy x = 7; x = - 2/3 là nghiệm của pt.
\(2,x^3+3x^2-\left(x+3\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}}\)
Vậy x = - 3; x = 1; x = - 1 là nghiệm của pt.
\(3,15x-5+6x^2-2x=0\)
\(\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{2}\end{cases}}}\)
Vậy x = 1/3; x = - 5/2 là nghiệm của pt.
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)
\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)
Vậy pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)
c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)
Trả lời:
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\)
\(\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)
\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)
\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)
Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)
nên pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)
\(\Leftrightarrow3x.\left(-9\right).2x=0\)
\(\Leftrightarrow-54x^2=0\)
\(\Leftrightarrow x^2=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0 là nghiệm của pt.
c, \(7-9x+2x^2=0\)
\(\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)
Vậy x = 7/2; x = 1 là nghiệm của pt.
d, trùng ý c
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)