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- 6x2+13x+6
=6x2+9x+4x+6
=3x(2x+3)+2(2x+3)
=(2x+3)(3x+2)
2. 6x2-15x+6
=6x2-12x-3x+6
=6x(x-2)-3(x-2)
=(x-2).3(2x-1)
3. 8x2 -2x-3
= 8x2-6x+4x-3
=2x(4x-3)+(4x-3)
=(4x-3)(2x+1)
4. 8x2-10x-3
=8x2+12x-2x-3
=4x(2x+3)-(2x+3)
=(2x+3)(4x-1)
5. -10x2+4x+6
=-10x2+10x-6x+6
=-10x(x-1)-6(x-1)
=(x-1).(-2)(5x+3)
6. 10x2-28x-6
=10x2-30x+2x-6
=10x(x-3)+2(x-3)
=(x-3).2(5x+1)
6x2 + 13x + 6 = 6x2 + 9x + 4x + 6 = 3x( 2x + 3 ) + 2( 2x + 3 ) = ( 2x + 3 )( 3x + 2 )
6x2 - 15x + 6 = 6x2 - 12x - 3x + 6 = 6x( x - 2 ) - 3( x - 2 ) = 3( x - 2 )( 2x - 1 )
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5+y\right)\left(x+5-y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(x^2+2\right)\left(5x-7\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5-y\right)\left(x+5+y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(5x-7\right)\left(x^2+2\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
=\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)
đặt \(x^2+7x+10=a\)
ta có \(a\left(a+2\right)-24=a^2+2a-24\)
\(=a^2+2a+1-25\)
\(=\left(a+1\right)^2-5^2\)
\(=\left(a+1-5\right)\left(a+1+5\right)\)
\(=\left(a-4\right)\left(a+6\right)\)
\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
a) = (x +5)2 - 22 = (x+5 -2)(x+5 +2) = (x+3)(x+7)
b) = x(x2 -1) -6(x-1)= x(x+1)(x-1) -6(x-1) = (x-1)(x(x+1)-6)
Xin lỗi bạn,mk ms học đến phân tích đa thức thành nhân tử nhóm nhiều hạng tử,còn phần này mk ms học còn yếu lắm.
1. \(-10x^2+11x+6\)
\(=-10x^2+15x-4x+6\)
\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)
\(=\left(-5x-2\right)\left(2x-3\right)\)
2.\(10x^2-4x-6\)
\(=2\left(5x^2-2x-3\right)\)
\(=2\left(5x^2+3x-5x-3\right)\)
\(=2\left[x\left(5x+3\right)-\left(5x+3\right)\right]\)
\(=2\left(x-1\right)\left(5x+3\right)\)
3. \(10x^2+7x-6\)
\(=10x^2+12x-5x-6\)
\(=2x\left(5x+6\right)-\left(5x+6\right)\)
\(=\left(2x-1\right)\left(5x+6\right)\)
4. \(10x^2-14x-12\)
\(=2\left(5x^2-7x-6\right)\)
\(=2\left(5x^2+3x-10x-6\right)\)
\(=2\left[x\left(5x+3\right)-2\left(5x+3\right)\right]\)
\(=2\left(x-2\right)\left(5x+3\right)\)