Phân tích \(x^2-26\) thành nhân tử?
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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4: \(\left(2x+3\right)^3-1\)
\(=\left(2x+3-1\right)\left(4x^2+12x+9+2x+3+1\right)\)
\(=\left(2x+2\right)\left(4x^2+14x+13\right)\)
\(=2\left(x+1\right)\left(4x^2+14x+13\right)\)
5: \(4x^2+20xy+25y^2=\left(2x+5y\right)^2\)
6: \(x^4-64xy^3\)
\(=x\left(x^3-64y^3\right)\)
\(=x\left(x-4y\right)\left(x^2+4xy+16y^2\right)\)
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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Bài 4:
\(x^3-2x^2+x=x\left(x-1\right)^2\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(x^2-12x+36=\left(x-6\right)^2\)
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\(36^2+26^2-52\cdot36\)
= \(36^2-2\cdot36\cdot26+26^2\)
= \(\left(36-26\right)^2\)
= \(10^2\)=\(100\)
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Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
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Phương trình 5 x 2 + 21x − 36 = 0 có a + b + c = 5 +21 – 26 = 0 nên phương trình có hai nghiệm phân biệt là x 1 = 1 ; x 2 = - 26 5 . Khi đó B = 5. (x − 1) x + 26 5
Đáp án: C
Ta có \(x^2-26\) thành nhân tử
\(x^2-26=x^2-\left(\sqrt{26}^2\right)=\left(x-\sqrt{26}\right)\left(x+\sqrt{26}\right)\)