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Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+7x+13\right)=0\Rightarrow x=-2\)
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow x^3+9x^2+27x+27=1\)
\(\Leftrightarrow\left(x+3\right)^3=1^3\)
\(\Leftrightarrow x+3=1\Leftrightarrow x=-2\)
a)
(x+4)(3x-5) = 0
=> x + 4 = 0 hoặc 3x-5 = 0
x = -4 x = 5/3
b)
2x2 + 7x + 3 = 0
2x2 + 6x + x + 3= 0
(2x+1)(x+3) = 0
=> 2x+1 = 0 hoặc x + 3 = 0
x = -1/2 x = -3
\(x^3+2x^2+3x=0\)\(\Leftrightarrow x.\frac{x^3+2x^2+3x}{x}=0\)
\(\Leftrightarrow x\left(x^2+2x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+2x+3=0\end{cases}}\)
Ta sẽ c/m \(x^2+2x+3=0\) vô nghiệm.Thật vậy:
\(x^2+2x+3=\left(x+1\right)^2+2\ge2\forall x\)
Từ đó suy ra \(x^2+2x+3=0\) vô nghiệm.
Vậy : x = 0
\(\left(x+2\right)\left(2x-1\right)+1=4x^2\)
\(2x^2-x+4x-2+1=4x^2\)
\(\Rightarrow2x^2-3x+1=0\)
\(2x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
ý còn lại tham khảo bài tth
x3 _ x2 _ 4x - 4 = 0
x mũ 2(x+1)- 4(x+1)=0
(x mũ 2 - 4) (x+1)=0
(x+2) (x-2) (x+1) =0
suy ra (x+2)=0
(x-2)=0
(x+1)=0
vậy x=-2
x=2
x= -1
good luck!
Sửa đề : \(x^3-x^2-4x+4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow x=\pm2;1\)