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a) \(x^2\)\(-5x+6\)
=\(x^2\)\(-3x-2x+6\)
=\(x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-2\right)\left(x-3\right)\)
b) \(3x^2\)\(+9x-30\)
=\(3x^2\)\(-6x+15x-30\)
=\(3x\left(x-2\right)+15\left(x-2\right)\)
=\(\left(x-2\right)\left(3x+15\right)\)
c)\(x^2\)\(-3x+2\)
=\(x^2\)\(-2x-x+2\)
=\(x\left(x-2\right)-\left(x-2\right)\)
=\(\left(x-2\right)\left(x-1\right)\)
d) \(12x^2\)\(+7x-12\)
=\(12x^2\)\(-9x+16x-12\)
=\(3x\left(4x-3\right)+4\left(4x-3\right)\)
=\(\left(3x+4\right)\left(4x-3\right)\)
e) \(15x^2\)\(+7x-2\)
=\(15x^2\)\(-3x+10x-2\)
=\(3x\left(5x-1\right)+2\left(5x-1\right)\)
=\(\left(3x+2\right)\left(5x-1\right)\)
f) \(a^2\)\(-5a-14\)
=\(a^2\)\(-7a+2a-14\)
=\(a\left(a-7\right)+2\left(a-7\right)\)
=\(\left(a+2\right)\left(a-7\right)\)
g) \(x^2\)\(-\left(a+b\right)x+ab\)
=\(x^2\)\(-ax-bx+ab\)
=\(x\left(x-a\right)-b\left(x-a\right)\)
=\(\left(x-a\right)\left(x-b\right)\)
a) \(A_4=\left(x^2-3x+5\right)^2+7x\cdot\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)^2+4x\cdot\left(x^2-3x+5\right)+3x\left(x^2-3x+5\right)+12x^2\)
\(=\left(x^2-3x+5\right)\left(x^2-3x+5+4x\right)+3x\left(x^2-3x+5+4x\right)\)
\(=\left[\left(x^2-3x+5\right)+3x\right]\cdot\left(x^2-3x+5+4x\right)\)
\(=\left(x^2-3x+5+3x\right)\left(x^2+x+5\right)\)
\(=\left(x^2+5\right)\left(x^2+x+5\right)\)
\(A_5=2\left(x^2+5x-2\right)^2-7\left(x^2+5x-2\right)\left(x^3+3\right)+5\left(x^2+3\right)^2\)
Đặt \(x^2+5x-2=a;x^3+3=b\),Ta có:
\(2a^2-7ab+5b^2=2a^2-5ab-2ab+5b^2=a\left(2a-5b\right)-b\left(2a-5b\right)=\left(2a+5b\right)\left(a-b\right)\)
Thay \(x^2+5x-2=a;x^3+3=b\),ta có:
.......................
bn làm nốt nhé
\(8x^3+12x^2+6x+7-3\left(2x+1\right)^2=6\)
\(\left(2x\right)^3+3\times\left(2x\right)^2\times1+3\times2x\times1^2+1^3+6-3\left(2x+1\right)^2=6\)
\(\left(2x+1\right)^3-3\left(2x+1\right)^2=6-6\)
\(\left(2x+1\right)^2\left(2x+1-3\right)=0\)
\(\left(2x+1\right)^2\left(2x-2\right)=0\)
\(2\left(2x+1\right)^2\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}2x+1=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=1\end{array}\right.\)
\(20x^3-15x^2+7x=45x^2-38x\)
\(20x^3-15x^2-45x^2+7x+38x=0\)
\(20x^3-60x^2+45x=0\)
\(5x\left(4x^2-12x+9\right)=0\)
\(5x\left(2x-3\right)^2=0\)
\(\left[\begin{array}{nghiempt}x=0\\2x-3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)
\(b,3x^2+7x+2=0\\ \Leftrightarrow3x^2+x+6x+2=0\\ \Leftrightarrow x\left(3x+1\right)+2\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
\(c,4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2\Leftrightarrow2x-3=0\\ \Leftrightarrow x=\dfrac{3}{2}\)
\(d,x^2-10x-2000=0\\ \Leftrightarrow\left(x-50\right)\left(x+40\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
ý a, hình như sai đề bn kiểm tra lại ik nha!
a) \(=7x^2+49x+x+7=7x\left(x+7\right)+\left(x+7\right)=\left(x+7\right)\left(7x+1\right)\)
c) \(=15x^2+10x-3x-2=5x\left(3x+2\right)-\left(3x+2\right)=\left(3x+2\right)\left(5x-1\right)\)
ta có : 7x2 + 49x + x + 7
= 7x(x + 7) + (x + 7)
= (x + 7) (7x + 1)
k mk mk k lại