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\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
\(=\left(x+\frac{1}{2}\right)\left(x-1\right)\left(x-\frac{2}{3}\right):\left(x+\frac{1}{2}\right)\)
\(=\left(x-1\right)\left(x-\frac{2}{3}\right)\)
a) Ta có: 6x^3 - 7x2 - x+2 = 6x3+3x2-10x2-5x+4x+2
= 3x2 ( 2x+1) - 5x(2x+1) + 2(2x+1)
= (2x+1)(3x2-5x+2)
Suy ra: (6x3-7x2-x+2): (2x+1)= 3x2-5x+2
b) Ta có: x2 -y2+6x+9= (x2+6x+9) - y2
= (x+3)2 - y2
= (x+3-y)(x+3+y)
Suy ra: (x2-y2+6x+9): (x+y+3)= x-y+3
Làm vậy nha pạn :)
Giải:
a) 6x - 7x - x + 2 2x + 1 3 2 3x - 5x + 2 2 6x + 3x 3 2 - -10x - x + 2 2 -10x - 5x 2 4x + 2 - 4x + 2 0
Vậy ...
b)
2x - 10x - x + 15x - 3 2x - 3 4 3 2 2 x - 5x +3/2 2 2x - 3x 4 - -10x + 3x + 15x - 3 2 3 2 -10x + 15x - 3 3x - 3 2 3x - 2 2 - -1
Vậy ...
c)
Vậy ...
Chúc bạn học tốt!
Ukm
It's very hard
l can't do it
Sorry!
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt
b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)
Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)
\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)
Từ đó tính đc x
d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+5=a\), khi đó pt có dạng:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
a) (x3 + 8y3) : (2y + x)
= (x + 2y)(x2 - 2xy + 4y2) : (2y + x)
= x2 - 2xy + 4y2
b) (x3 + 3x2y + 3xy2 + y3) : (2x + 2y)
= (x + y)3 : 2(x + y)
= \(\dfrac{\left(x+y\right)^2}{2}\)
c) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= 3x3y2(2x2 - 3xy + 5y2) : 3x3y2
= 2x2 - 3xy + 5y2
(x2– y2 + 6x + 9) : (x + y + 3) = (x2 + 6x+ 9) – y2 : (x + y + 3)
=(x + 3)2 – y2 : (x + y + 3) = (x + 3 – y) (x + 3 + y) : (x + y + 3) = (x – y + 3)
còn câu c thì sao bn