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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 \(A⋮B\)
a) \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left(2x-3\right)^2\)
b)\(1-9x+27x^2-27x^3\)
\(=\left(1-3x\right)^{^3}\)
c)\(\frac{x^2}{4}+2xy+4y^2-25\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
d)\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)^2-3\left(x^2-4x\right)-5\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)\left(x^2-4x-3\right)-5\left(x^2-4x-3\right)\)
\(=\left(x^2-4x-5\right)\left(x^2-4x-3\right)\)
\(=\left(x^2+x-5x-5\right)\left(x^2-4x-3\right)\)
\(=\left[x\left(x+1\right)-5\left(x+1\right)\right]\left(x^2-4x-3\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x^2-4x-3\right)\)
Chúc bạn học tốt !
bài 1
a(x+y)2-(x-y)2
=[(x+y)-(x-y)][(x+y)+(x-y)]
=(x+y-x+y)(x+y+x-y)
=2y.2x
b,(3x+1)2-(x+1)2
=[(3x+1)-(x+1)][(3x+1)+(x+1)]
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
4x.(x+10
bài 2
x3-0,25x=0
=>x(x2-0,25)=0
=>x=0 hoặc x2-0,25=0
=> x=0 hoặc x=\(\pm0,5\)
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
A/ \(2x^2+7x+5=2\left(x^2+2x+1\right)+3x+3=2\left(x+1\right)^2+3\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
B/ \(x^2-4x-5=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)
C/ \(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
D/\(x^4+4x^2-5=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-3^2=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) = 2x^2 + 2x +5x + 5 = 2x(x+1) + 5(x+1) = (2x+5)(x+1)
b) = x^2 + x - 5x - 5 = x(x-1) - 5(x-1) = (x-5)(x-1)
c) = x^3 ( x+1) + x+1 = (x^3+1) (x+1) = (x+1)^2 * (x^2 - x +1)
d) = x^4 - x^2 + 5x^2 -5 = x^2 (x^2-1) + 5(x^2-1) = (x^2+5)(x-1)(x+1)
Ta có ; x2 - 11x + 24
= x2 - 3x - 8x + 24
= x(x - 3) - (8x - 24)
= x(x - 3) - 8(x - 3)
= (x - 3)(x - 8)
a) \(x^2+4x-y^2+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
Bài 1:
a: \(49-4x^2=\left(7-2x\right)\left(7+2x\right)\)
b: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
c: \(x^2+18xy+81y^2=\left(x+9y\right)^2\)
bn ơi có bài 2,3 ko?
?