phân tích thành nhân tử \(8x^2+30x+7\)
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a)
ta có \(x^2+8x+7\)
\(=x^2+7x+x+7\)
\(=x\left(x+7\right)+\left(x+7\right)\)
\(=\left(x+7\right)\left(x+1\right)\)
b)
Ta có \(8x^2+30x+7\)
\(=8x^2+2x+28x+7\)
\(=2x\left(4x+1\right)+7\left(4x+1\right)\)
\(\left(4x+1\right)\left(2x+7\right)\)
a) \(=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)
b) \(=x\left(x+y\right)+8\left(x+y\right)=\left(x+y\right)\left(x+8\right)\)
c) \(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
a) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+2x+28x+7=0\)
\(\Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(4x+1\right)=0\)
\(\Rightarrow\)\(2x+7=0\) hoặc \(4x+1=0\)
\(\Rightarrow\)\(2x=-7\) ; \(4x=-1\)
\(\Rightarrow\)\(x=\frac{-7}{2}\) ; \(x=\frac{-1}{4}\)
Vậy \(x\in\left\{\frac{-7}{2};\frac{-1}{4}\right\}\)
b) \(x^3-11x^2+30x=0\)
\(\Rightarrow x\left(x^2-11x+30\right)=0\)
\(\Rightarrow x\left(x^2-6x-5x+30\right)=0\)
\(\Rightarrow x\left[x\left(x-6\right)-5\left(x-6\right)\right]=0\)
\(\Rightarrow x\left(x-5\right)\left(x-6\right)=0\)
\(\Rightarrow\)\(x=0\) hoặc \(x-5=0\) hoặc \(x-6=0\)
\(\Rightarrow\)\(x=0\) ; \(x=5\) ; \(x=6\)
Vậy \(x\in\left\{0;5;6\right\}\)
a)\(8x^2+30x+7=0\Leftrightarrow8x^2+2x+28x+7=0\Leftrightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Leftrightarrow\left(2x+7\right)\left(4x+1\right)=0\Leftrightarrow\orbr{\begin{cases}2x+7=0\\4x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
b)\(x^3-11x^2+30x=0\Leftrightarrow x\left(x^2-11x+30\right)=0\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)
\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\Leftrightarrow x\left(x-6\right)\left(x-5\right)=0\)
<=>x=0 hoặc x-6=0 hoặc x-5=0 <=> x=0 hoặc x=6 hoặc x=5
( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
\(2x^4-8x^3-7x^3+28x^2+7x^2-28x-2x+8\\ =2x^3\left(x-4\right)-7x^2\left(x-4\right)+7x\left(x-4\right)-2\left(x-4\right)\\ =\left(x-4\right)\left(2x^3-7x^2+7x-2\right)\\ =\left(x-4\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)\\ =\left(x-4\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]\\ =\left(x-4\right)\left(x-2\right)\left(2x^2-2x-x+1\right)\\ =\left(x-4\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)\)
\(-\left(x+2y\right)^2\)
\(-\left(x-3\right)^2\)
\(\left(3-5x\right)^2\)
\(-x^2-4xy-4y^2=-\left(x+2y\right)^2\)
\(-x^2+6x-9=-\left(x-3\right)^2\)
\(25x^2-30x+9=\left(5x-3\right)^2\)
\(x^2-y^2+8x+6y+7\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)
\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)
\(=\left(x+y+1\right)\left(x-y+7\right)\)
8x2+30x+7
=8x2+2x+28x+7
=(8x2+2x)+(28x+7)
=2x(4x+1)+7(4x+1)
=(2x+7)(4x+1)