Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) kết quả là x^2-2x+3
b) CM NÈ:
X^2-2X+3=(X^2-2X+1)+2=(X-1)^2+2
VÌ (X-1)^2>=0 VỚI MỌI X=>(X-1)^2+2>0 VỚI MỌI x=> GIÁ TRỊ BIỂU THỨC LUÔN DƯƠNG
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2. (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
Bài làm :
1) (x2 - 2x - 1)(x - 3)
= x2(x - 3) - 2x(x - 3) - 1(x - 3)
= x3 - 3x2 - 2x2 + 6x - x + 3
= x3 - 5x2 + 5x + 3
2) (-x + 4)(-x2 + 4x - 1)
= -x(-x2 + 4x - 1) + 4(-x2 + 4x - 1)
= x3 - 4x2 + x - 4x2 + 16x - 4
= x3 - 8x2 + 17x - 4
3 ) (2x - 1)(x2 - 5x + 3)
= 2x(x2 - 5x + 3) - 1(x2 - 5x + 3)
= 2x3 - 10x2 + 6x - x2 + 5x - 3
= 2x3 - 11x2 + 11x - 3
a: \(x^3-x^2-14x+24\)
\(=x^3+x-12-x^2-15x+36\)
=>\(\left(x^3-x^2-14x+24\right):\left(x^3+x-12\right)=1+\frac{-x^2-15x+36}{x^3+x-12}\)
Để dư là 0 thì \(-x^2-15x+36=0\)
=>\(x^2+15x-36=0\) (1)
\(\Delta=15^2-4\cdot1\cdot\left(-36\right)=225+144=369>0\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{-15-\sqrt{369}}{2\cdot1}=\frac{-15-3\sqrt{41}}{2}\\ x=\frac{-15+3\sqrt{41}}{2}\end{array}\right.\)
b: \(x^5+4x^3+3x^2-5x+15\)
\(=x^5-x^3+3x^2+5x^3-5x+15=\left(x^3-x+3\right)\left(x^2+5\right)\)
=>\(\frac{x^5+4x^3+3x^2-5x+15}{x^3-x+3}=x^2+5\)
=>Đây là phép chia hết
c: \(2x^4+2x^3+3x^2-5x-20\)
\(=2x^4+2x^3+8x^2-5x^2-5x-20=\left(x^2+x+4\right)\left(2x^2-5\right)\)
=>\(\frac{2x^4+2x^3+3x^2-5x-20}{x^2+x+4}=2x^2-5\)
d: \(2x^4-14x^3+19x^2-20x+9\)
\(=2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\)
\(=\left(x^2-4x+1\right)\left(2x^2-6x-7\right)-42x+16\)
=>\(\frac{2x^4-14x^3+19x^2-20x+9}{x^2-4x+1}=2x^2-6x-7\) dư -42x+16
để dư bằng 0 thì -42x+16=0
=>-42x=-16
=>\(x=\frac{16}{42}=\frac{8}{21}\)
4x^3-3x^2 +1 x^2+2x-1 4x 4x^3+8x^2-4x - -11x^2+4x+1 -11 -11x^2-22x+11 - 26x-10
OLM chỉ có phần chụp ảnh cho CTV
Lưu ý bạn cố phải viết thẳng hàng vì OLM ko viết đc
a: \(\left(5x-3\right)\left(5x+3\right)=25x^2-9\)
b: \(=2x^2-12x^3+15x^2-18x=-12x^3+17x^2-18x\)
c: \(\dfrac{16x^3y^4}{4x^2y^2}=4xy^2\)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)