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Làm trc cho 2 câu cuối
c) \(a^2-b^2-4a+4b\)
\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)
\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)
d) \(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)+1\)
\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)
\(\left(x+2\right)\left(x^2+2x-9\right)\)
\(=x^3+2x^2-9x+2x^2+4x-18\)
\(=x^3+4x^2-5x-18\)
\(\left(x^{2y}-6\right)\left(x^2-5\right)\)
\(=x^{4y}-5x^{2y}-6x^2+30\)
\(\left(x+y\right)\left(xy-4+y\right)\)
\(=x^2y-4x+xy+xy^2-4y+y^2\)
câu còn lại tương tự nha
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
2: \(N=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{39}{2}+1-3-2=\dfrac{39}{2}-4=\dfrac{31}{2}\)
3: \(P=4x^2-25-4x^2-4x-1=-4x-26\)
=-8020-26=-8046
4: \(Q=\left(y^2-9\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=y^4-81-y^4+4=-77\)
a, \(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(=-\left(1^2-2^2+3^2-4^2+...+99^2-100^2\right)\)
\(=-\left[\left(1+2\right)\left(1-2\right)+\left(3+4\right)\left(3-4\right)+...+\left(99+100\right)\left(99-100\right)\right]\)
\(=-\left(-3-7-...-199\right)\)
\(=3+7+...+199\)
\(=\frac{\left(199+3\right).50}{2}=5050\)