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\(x^2+7x-a^2+a+12=0\)
\(\Leftrightarrow x^2-ax+4x+ax+3x-a^2+a+12=0\)
\(\Leftrightarrow\left(x^2-ax+4x\right)+\left(ax+3x\right)-\left(a^2+3a\right)+\left(4a+12\right)=0\)
\(\Leftrightarrow x\left(x-a+4\right)+x\left(a+3\right)-a\left(a+3\right)+4\left(a+3\right)=0\)
\(\Leftrightarrow x\left(x-a+4\right)+\left(a+3\right)\left(x-a+4\right)=0\)
\(\Leftrightarrow\left(x+a+3\right)\left(x-a+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+a+3=0\\x-a+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-a-3\\x=a-4\end{cases}}}\)
Vậy \(x=-a-3\) hoặc \(x=a-4\)
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Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
![](https://rs.olm.vn/images/avt/0.png?1311)
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
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Ta có T = (x + 2)(x + 3)(x + 4)(x + 5) – 24
= [(x + 2)(x + 5)].[(x + 3)(x + 4)] – 24
= ( x 2 + 7x + 10).( x 2 + 7x + 12) – 24
Đặt x 2 + 7x + 11= t, ta được
T = (t – 1)(t + 1) – 24 = t 2 – 1 – 24 = t 2 – 25 = (t – 5)(t + 5)
Thay t = x 2 + 7x + 11, ta được
T = (t – 5)(t + 5) = ( x 2 + 7x + 11 – 5)( x 2 + 7x + 11 + 5)
= ( x 2 + 7x + 6)( x 2 + 7x + 16)
Suy ra a = 6; b = 16 => a – b = -10
Đáp án cần chọn là: D
![](https://rs.olm.vn/images/avt/0.png?1311)
a)x(x+3)+a(x-3)=2(ax-1)
=>x2+3x+ax-3a=2ax-2
=>x2+3x+ax-3a-2ax=-2
=>x2+3x-ax-3a = -2
=> (x2+3x)-(ax+3a)=-2
=>x(x+3)-a(x+3)=-2
=>(x+3)(x-a)=-2
=>x+3và x-a\(\in\)U(-2)
x+3=>x | x-a=>a | |
-2 | x=-5 |
a=-6 |
-1 | x=-4 | a=-6 |
1 | x=-2 | a=0 |
-2 | x=-1 | a=-3 |
vậy S={-5;-4;-2;-1}lần lượt tương ứng với a\(\in\){-6(hai lân);0;-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow7x\left(x+5\right)+\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(7x+x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(8x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{5}{8}\end{matrix}\right.\)
\(\Leftrightarrow x^2+x-5x-5-x^2-6x-9-6=0\\ \Leftrightarrow-10x-20=0\\ \Leftrightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
=(x2+7x+12)-a2+a=0
=(x+3)(x+4)-a2+a=0
Đặt x+3=y.Ta có
=y2+y-a2+a=0
=(y-a+1)(y+a)=0
=>y=a-1,y=-a
mà x+3=y
=>x=a-4 và x=-a-3