a) (x^2 + x )^2 + 4(x^2 + x ) = 12
b) 6x^4 - 5x^3 - 38x^2 - 5x + 6 = 0
c) Tìm GTNN:
x^2 + 5y^2 + 2xy - 4x - 8y + 2015
Giải pT sau : a.x(4x-1)^2(2x-1)=9 b.(x^2+5x+6)(x^2-11x+30)=180 c.6x^4-5x^3-38x^2-5x+6=0
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
tìm gtnn của
a)A=x^2-3x
b)B=2x^2-x
c)C=5x^2+4y-4xy-4x
d)D=x^2+5y-4xy-6x+8y+12
Bạn cũng cần xem lại đề câu c nhé.
Tìm GTNN của các biểu thức sau:
a,A=x^2-6x+15
b,B=x^2+4x
c,C=2x^2-x+4
d,D=x^2+x+1
e,E=2x^2-5x-2
f,F=x^2+5y^2+2xy-y+3
Mk chỉ làm hai bài đầu gợi ý thôi chứ mk cũng ko đủ TG
a)\(A=x^2-6x+15\)
\(\Leftrightarrow A=x^2-6x+9+6\)
\(\Leftrightarrow A=\left(x-3\right)^2+6\)
Vì \(\left(x-3\right)^2\ge0\)\(\Rightarrow\)\(\left(x-3\right)^2+6\ge6\)
Dấu = xảy ra khi x - 3 = 0 ; x = 3
Vậy Min A = 6 khi x=3
b)\(B=x^2+4x\)
\(\Leftrightarrow B=x^2+4x+4-4\)
\(\Leftrightarrow B=\left(x+2\right)^2-4\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2-4\ge-4\)\
Dấu = xảy ra khi x + 2 = 0 ; x = -2
Vậy Min B = -4 khi x =-2
Giải phương trình:
a, (x^2+x)^2+4(x^2+x)=12
b, 6x^4-5x^3-38x^2-5x+6=0 (phương trình có hệ số đối xứng bậc 4)
\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)
\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=-2\\3x=1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\\x=\dfrac{1}{3}\\x=3\end{matrix}\right.\\ \text{Vậy }S=\left\{-\dfrac{1}{2};-2;\dfrac{1}{3};3\right\}\)
Phân tích đa thức sau thành nhân tử :
a, -x - y^2 + x^2 - y
b, x( x + y ) - 5x - 5y
c,x^2 - 5x + 5y - y^2
d, 5x^3 - 5x^2 y - 10x^2 + 10xy
e,27x^3 - 8y^3
f, x^2 - y^2 - x - y
g, x^2 - y^2 - 2xy + y^2
h, x^2 - y^2 + 4 - 4x
i, x^6 - y^6
a) x^2+2xy+y^2-16
b) 3x^2+5x-3xy-5y
c) 4x^2-6x^3y-2x^2+8x
d) x^2-4-2xy+y^2
e) x^3-4x^2-12x+27
g) 3x^2-18x+27
h) x^2-y^2-z^2-2yz
k) 4x^2(x-6)+9y^2(6-x)
l)6xy+5x-5y-3x^2-3y^2
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
thu gọn đa thức
a) \(4x^5y^2-9x^2y^4+3x^5y^2+5x^2y^4-6x^6\)
b) \(5x^8y^2-x^2y+3x^2y-5x^8y^2+6x^2y\)
Bài 1: Viết thành tích
1: 81-100n^8.
2:m^2 -25
3:16x^2 -25y^2
4: 100a^2b^4 -49
5:8x^3 -27y^6
6: 64 + 125a^6
7: x^6 -y^6
8: -8y^9 -x^6
9:1/4x^4y^2 -3x^5y +9x^6
10:x^2 -2xy -16 +y^2
11: x^2 -6x -y^2 +9
12:x^2 -9 +4xy + 4y^2
Bài 2: Phân tích đa thức thành nhân tử
a, x^4 +2x^3 +x^2
b, x^3 -x +3x^2y + 3xy^2 +y^3 -y
c, 5x^2 -10 xy +5y^2 -20 z^2
d, x^2 +5x -6
e, 5x^2+5xy - x - y
f, 7x -6x^2 -2
g, x^2 +4x +3
h, 2x^2 +3x -5
i, 16x - 5x^2 -3
j, (x^2 +x)^2 +1 (x^2 +x) -12
k, (x^2 +x +1) . (x^2 +x +1) -12
I, (x+1) . (x+2) .(x+3) .(x+4) -24
Bài3:Tìm x biết
1, (x+3) (x^2 -3x +9) - x(x^2-3) = 8(5 -x)
2, (2x +1)^3 +(2x+3)^3 = 0
3, (5x +1)^2 - (5x -3) (5x +3) = 30
4, ( x+3) (x^2 -3x+9) - x(x-2) (x+2) = 15
Mọi người làm giúp mình vs mai mình nộp r
phân tích đa thức
a)x^4+6x^3+11x^2+6x+1
b)x^4+x^3+x^2+x+1
c)6x^4+5x^3-38x^2+5x+6
d)x^4+5x^3-12x^2+5x+1
dễ mà bạn xin 20 phút làm ra giấy nhé :))
a) \(\left(x^4+6x^3+9x^2\right)+2x^2+6x+1\)
\(\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(\left(x^2+3x+1\right)^2\)
b) \(x^4+x^3+x^2+x+1\)
câu b, chúa sẽ c/m x ko tồn tại , và nó là 1 đa thức bất khả Q . trong R
vì lớp 8 chưa học đến số phức
\(x^4+x^3=-x^2-x-1\)
\(x^4+x^3+\frac{1}{4}x^2=\left(\frac{1}{4}x^2-x^2\right)-x-1\)
\(\left(x^2+\frac{1}{2}x\right)^2=-\frac{3}{4}x^2-x-1\)
\(4\left(x^2+\frac{1}{2}x\right)^2=-3x^2-4x-4\)
\(\Delta`=\left(-2\right)^2-\left(-4\right).\left(-3\right)=4-12< 0\)
denta < 0 x vô nghiệm
vậy đa thức trên ko thể phân tích và nó là 1 đa thức bất khả Q
c) ,
\(\left(6x^4-12x^3\right)+\left(17x^3-34x^2\right)-\left(4x^2-8x\right)-\left(3x-6\right)\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)\)
\(\left(x-2\right)\left\{\left(6x^3+18x^2\right)-\left(x^2+3x\right)-\left(x+3\right)\right\}\)
\(\left(x-2\right)\left\{6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right\}\)
\(\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)\)
\(\left(x-2\right)\left(x+3\right)\left\{\left(6x^2+\frac{6}{3}x\right)-\left(\frac{9}{3}x+\frac{9}{9}\right)\right\}\)
\(\left(x-2\right)\left(x+3\right)\left\{6x\left(x+\frac{1}{3}\right)-\frac{9}{3}\left(x+\frac{1}{3}\right)\right\}\)
\(\left(X-2\right)\left(X+3\right)\left(X+\frac{1}{3}\right)\left(6x-1\right)\)
d)
\(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)\)
\(x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)
\(\left(x-1\right)\left(x^3+6x^2-6x-1\right)\)
\(\left(x-1\right)\left\{\left(x^3-x^2\right)+\left(7x^2-7x\right)+\left(x-1\right)\right\}\)
\(\left(x-1\right)^2\left(x^2+7x+1\right)\)
\(\Delta=49-4=45\)
\(x1,2=\frac{-7+\sqrt{45}}{2},\frac{-7-\sqrt{45}}{2}\)
\(\left(x-1\right)^2\left(x-\frac{7+\sqrt{45}}{2}\right)\left(x-\frac{7-\sqrt{45}}{2}\right)\)