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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 1
2x2 + 8x + 16 = 2(x2 + 4x + 4) = 2(x + 2)2
Bài 2
\(\frac{x}{x-5}\)\(+\)\(\frac{2}{x^2-25}\)\(=\)\(\frac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)\(+\)\(\frac{2}{x^2-25}\)\(=\)\(\frac{x^2+5x+2}{x^2-25}\)
\(\frac{x}{x-5}+\frac{2}{x^2-25}=\frac{x\left(x+5\right)+2}{x^2-25}\)
\(=\frac{x^2+5x+2}{x^2-25}\)
a) x4 + 4 = (x4 + 4x2 + 4) - 4x2 = (x2 + 2)2 - 4x2 = (x2 + 2x + 2)(x2 - 2x + 2)
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24 = (x + 2)(x + 5)(x + 3)(x + 4) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = y => y(y + 2) - 24 = y2 + 2y - 24
= y2 + 6y - 4y - 24 = (y - 4)(y + 6) = (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16) = (x2 + x + 6x + 6)(x2 + 7x + 16) = (x + 1)(x + 6)(x2 + 7x + 16)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
k cho tui nha
\(a,x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
\(b,25-4x^2-4xy-y^2\)
\(=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x+y\right)\)
\(c,x^3-x+y^3-y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+1\right)\)
a)\(6x^2-9xy\)
\(=3x\left(2x-3y\right)\)
b)\(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
c)\(x^4-8x^2-9\)
\(=x^4+x^2-9x^2-9\)
\(=x^2\left(x^2+1\right)-9\left(x^2+1\right)\)
\(=\left(x^2-9\right)\left(x^2+1\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(x^2+1\right)\)
d)\(x^4-4\left(x^2+5\right)-25\)
\(=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-5-4\right)\)
\(=\left(x^2+5\right)\left(x^2-9\right)\)
\(=\left(x^2+5\right)\left(x-3\right)\left(x+3\right)\)
b) 4x2 - 25 + (2x + 5)2
= (2x + 5)(2x - 5) + (2x + 5)2
= (2x + 5)(2x - 5 + 2x + 5)
= 4x(2x + 5)