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\(\left(m^3-m+1\right)^2+\left(m^2-3\right)-2\left(m^2-3\right)\left(m^3-m+1\right)\)
\(=\left(m^3-m+1+m^2-3\right)^2\)
\(=\left(m^3+m^2-m-2\right)^2\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
Bài 2:
a: \(\Leftrightarrow4x^2-4x+1-4x^2-16x-16=9\)
=>-20x-15=9
=>-20x=24
=>x=-6/5
b: \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
=>9x=18
=>x=2
1/Đặt Q(x) là thương ta có
\(x^3-7x^2+a=Q\left(x\right).\left(x-2\right)\).Thay x=2 đc
\(8-28+a=0\Leftrightarrow a=20\)
2/a/ĐKXĐ: x khác 2,-3
Có \(M=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(\Leftrightarrow M=\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow M=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow M=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow M=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow M=\frac{x-4}{x-2}\)
b/\(M=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\).Để M nguyên thì \(2⋮x-2\Rightarrow x-2\in\left(+-1,+-2\right)\Rightarrow x\in\left(3,1,4,0\right)\)
\(\left(m^2-m+1\right)^2+\left(m^2-3\right)^2-2\left(m^2-3\right)\left(m^2-m+1\right)\)
\(=\left(m^2-m+1-m^2+3\right)^2\)
\(=\left(4-m\right)^2\)
Chúc bạn học tốt!!!
\(\left(m^n-m+1\right)^2+\left(m^2-3\right)^2-2\left(m^2-3\right)\left(m^2-m+1\right)\)
\(=\left(m^2-m+1-m^2+3\right)^2\)
\(=\left(4-m\right)^2\)