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\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
a) \(2x^2+5x-18\)
\(=2x^2-4x+9x-18\)
\(=2x\left(x-2\right)+9\left(x-2\right)\)
\(=\left(x-2\right)\left(2x+9\right)\)
b) \(4x^2-17x+15\)
\(=4x^2-12x-5x+15\)
\(=4x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(4x-5\right)\)
c) \(-8x^2+10x+7\)
\(=-8x^2-4x+14x+7\)
\(=-4x\left(2x+1\right)+7\left(2x+1\right)\)
\(=\left(2x+1\right)\left(-4x+7\right)\)
d) \(7x^2-30x+8\)
\(=7x^2-28x-2x+8\)
\(=7x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(7x-2\right)\)
e) \(-x^3+11x^2-30x\)
\(=x\left(-x^2+11x-30\right)\)
\(=x\left(-x^2+5x+6x-30\right)\)
\(=x\left[-x\left(x-5\right)+6\left(x-5\right)\right]\)
\(=x\left(x-5\right)\left(-x+6\right)\)
a) 2x\(^2\) + 5x - 18 = 2x\(^2\) + 9x - 4x - 18 = x(2x + 9) - 2(2x + 9) = (x-2)(2x-9)
b) 4x\(^2\) - 17x - 15 = 4x\(^2\) + 20x - 3x - 15 = 4x(x + 5 ) - 3(x + 5) = (4x - 3 )(x + 5)
c) -8x\(^2\) + 10x + 7 = -8x\(^2\) + 14x - 4x + 7 =-2x(4x - 7) - (4x - 7) = (-2x - 1)(4x - 7)
d) 7x\(^2\) - 30x + 8 = 7x\(^2\) + 2x + 28x + 8 = x(7x + 2) + 4(7x + 2) = (x + 4)(7x + 2)
e) - x\(^3\) + 11x\(^2\) - 30x = -x(x\(^2\) - 11x + 30) = -x(x\(^2\) - 5x - 6x + 30) = -x\(\left[x\left(x-5\right)-6\left(x-5\right)\right]\) = -x(x-6)(x-5)
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) ( a + b + c ) 2 + ( a + b - c ) 2 -2 x ( a+b) 2
2a+2b+2x+2a+2b-2c-2.(2a+2b)
2a+2b+2c+2a+2b-2c-4a-4b
4a+4b-4a-4b=0
b) 2x.( 2x -1 ) 2 -3x.( x+3 )( x-3) - 4x.(x+1).2
2x.(4x-2)-3x2-9x-3x2+9x-4x(2x+2)
8x2-4x-3x2-9x-3x2+9x-8x2-8x
-12x-3x2
c) ( a-b+c).2 -(b-c).2 + 2ab - 2ac
2a-2b+2c-2b+2c+2ab-2ac
2a-4b+4c+2ab-2ac
d) (3x+1).2 - 2(3x+1)( 3x+5 )+(3x+5).2
6x+2-6x-2-6x-10+6x+10=0
Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3
bài 1 : ta có : \(A=27x^3+27x^2y+9xy^2+y^3=\left(3x+y\right)^3\)
\(=\left(3.\left(-3\right)+5\right)^3=\left(-9+5\right)^3=\left(-4\right)^3=-64\)
bài 2 : a) ta có : \(12a^2-3ab+8ac-2bc=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(3a+2c\right)\left(4a-b\right)\) câu này mk sữa đề lại chút .
b) ta có : câu này đề sai rồi .
nếu phân tích ra nó sẽ thành : \(17x^2+34x-5=\left(17x+17-\sqrt{374}\right)\left(x+\dfrac{17+\sqrt{374}}{17}\right)\)
c) ta có : \(4x^4+81=\left(2x^2\right)^2+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
câu 3 : a) ta có : \(-3x^2+2x+1=0\Leftrightarrow-3x^2+3x-x+1=0\)
\(\Leftrightarrow-3x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(-3x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-1}{3};x=1\)
b) ta có : \(x\left(x-3\right)=2x-6=x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
vậy \(x=2;x=3\)