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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) \(\dfrac{x^4-2x^3}{2x^2-x^3}=\dfrac{x^3\left(x-2\right)}{x^2\left(2-x\right)}=\dfrac{-x^3}{x^2}=-x\)
Thay x vào ta có biểu thức đã cho bằng\(-\left(\dfrac{-1}{2}=\dfrac{1}{2}\right)\)
a) <=> (8-5x+x-2)(x+2) + 4(x^2-x-2)=0
<=> 6x +12 - 4x^2 - 8x +4x^2 -4x -8 =0
<=> -6x -4 = 0
<=> x= 4/6
Bài 1:(Theo mình câu a nên sửa lại như thế này nhé)
a, a2-5a-14 b,x4+x2-2
=a2+2a-7a-14 =x4-x3+x3-x2+2x2-2x+2x-2
=(a2+2a)-(7a+14) =(x4-x3)+(x3-x2)+(2x2-2x)+(2x-2)
=a(a+2)-7(a+2) =x3(x-1)+x2(x-1)+2x(x-1)+2(x-1)
=(a+2)(a-7) =(x-1)(x3+x2+2x+2)
=(x-1)[(x3+x2)+(2x+2)]
=(x-1)[x2(x+1)+2(x+1)]
=(x-1)(x+1)(x2+2)
Bài 2:
a, x3+x2+x+1=0
<=>(x3+x2)+(x+1)=0
<=>x2(x+1)+(x+1)=0
<=>(x+1)(x2+1)=0
<=>\(\orbr{\begin{cases}x+1=0\\x^2+1=0\left(loại\right)\end{cases}}\)(x2 luôn lớn hơn hoặc bằng 0 =>x2+1 luôn lớn hơn hoặc bằng 1 nên x2+1=0 loại nhé)
<=>x= -1
b, x(2x-7)-4x+14=0
<=>x(2x-7)-(4x-14)=0
<=>x(2x-7)-2(2x-7)=0
<=>(2x-7)(x-2)=0
<=>\(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm a\end{matrix}\right.\)
Ta có: \(\dfrac{4}{x+a}+\dfrac{8}{x-a}=\dfrac{5a^2}{x\left(x^2-a^2\right)}\)
\(\Leftrightarrow\dfrac{4x\left(x-a\right)}{x\left(x+a\right)\left(x-a\right)}+\dfrac{8x\left(x+a\right)}{x\left(x+a\right)\left(x-a\right)}=\dfrac{5a^2}{x\left(x-a\right)\left(x+a\right)}\)
Suy ra: \(4x^2-4xa+8x^2+8xa-5a^2=0\)
\(\Leftrightarrow12x^2+4xa-5a^2=0\)
\(\Leftrightarrow12x^2+10xa-6xa-5a^2=0\)
\(\Leftrightarrow2x\left(6x+5a\right)-a\left(6x+5a\right)=0\)
\(\Leftrightarrow\left(6x+5a\right)\left(2x-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+5a=0\\2x-a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-5a\\2x=a\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5a}{6}\\x=\dfrac{a}{2}\end{matrix}\right.\)