(x-3)2- 2(x-1)=x(x-2)2 - 5x2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
20 tháng 10 2021
a: \(5x^2\left(3x^3-2x^2+x+2\right)\)
\(=15x^5-10x^4+5x^3+10x^2\)
b: \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
27 tháng 10 2021
Bài 1:
\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3-9x^2y+xy^2-3y^3+5x^2y-15xy^2=3x^3-3y^3-4x^2y-14xy^2\)
Bài 2:
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=x^2+16x+64-2x^2-12x+32+x^2-4x+4=100\\ c,=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
T
2
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
20 tháng 3 2021
Bài 2:
\(A=\dfrac{2}{-x^2-2x-2}=\dfrac{-2\left(-x^2-2x-2\right)-2x^2-4x-2}{-x^2-2x-2}\) \(=-2+\dfrac{2\left(x+1\right)^2}{-x^2-2x-2}\ge-2\)
Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy \(A_{Min}=-2\) khi \(x=-1\)
20 tháng 3 2021
Bài 1:
a) Ta có: \(2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)
5 tháng 7 2023
A) -2x(3x+2)(3x-2)+5(x+2)2 - (x-1)(2x+1)(2x+1)
= -2x(9x2-4)+5(x2+4x+4) - (x-1)(4x2-1)
= -18x3+8x+5x2+20x+20-(4x3-x-4x2+1)
= -18x3+5x2+28x+20-4x3+x+4x2+1
= -22x3+9x2+29x+21
B) (7x-8)(7x+8)-10(2x+3)2+5x(3x-2)2-4x(x-5)2
= 49x2 - 64 -10(4x2+ 12x + 3) + 5x(9x2 - 12x +4) - 4x(x2 - 10x +25)
= 49x2 - 64 -40x2 - 120x - 30 + 45x3 - 60x2 - 20x - 4x3 + 40x2 -100x
= 41x3 -11x2 -240x -94
6 tháng 7 2023
C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)
\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)
\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)
\(-5x^4-x^3+5x^2+20x-9\)
D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(-40x^4+36x^3+82x^2+6x-11\)
5 tháng 10 2021
Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)
\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)
Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)
\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)
5 tháng 10 2021
Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)
\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)
Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)
\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\left(x-3\right)^2-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow\left(x^2-6x+9\right)-\left(2x-2\right)=x\left(x^2-4x+4\right)-5x^2\)
\(\Leftrightarrow x^2-6x+9-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow x^2-8x+11=x^3-9x^2+4x\)
Đưa về bậc 3 giải nốt
\(\left(x-3\right)^2-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow\left(x-3\right)^3-2\left(x-1\right)-x\left(x-2\right)^2+5x^2=0\)
\(\Rightarrow x^3-9x^2+27x-27-2x-x^3+4x^2-4x+5x^2=0\)
\(\Rightarrow21x=25\Leftrightarrow x=\frac{25}{21}\)