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4 tháng 2 2017
a) x3+4x2+x-6=0
<=> x3+x2-2x+3x2+3x-6=0
<=>x(x2+x-2)+3(x2+x-2)=0
<=>(x+3)(x2+x-2)=0
<=>(x+3)(x2+2x-x-2)=0
<=>(x+3)[x(x+2)-(x+2)]=0
<=>(x+3)(x-1)(x+2)=0
=> x+3=0 hay
x-1=0 hay
x+2=0
<=> x=-3 hay x=1 hay x=-2
4 tháng 2 2017
b)x3-3x2+4=0
\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left\{\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
5 tháng 11 2017
\(a,x^4-4x^3+x^2-4x=0\)
\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
\(b,x^3-5x^2+4x-20=0\)
\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
AN
5 tháng 11 2017
a) \(x^4-4x^3+x^2-4x=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=0; x=4
b) \(x^3-5x^2+4x-20=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)
Vậy x=5
YN
29 tháng 11 2021
Answer:
\(5x^2-10xy+5y^2-20z^2\)
\(=5.\left(x^2-2xy+y^2-4z^2\right)\)
\(=5.[\left(x+y\right)^2-\left(2z\right)^2]\)
\(=5.\left(x+y-2z\right).\left(x+y+2z\right)\)
\(16x-5x^2-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(1-5x\right).\left(x-3\right)\)
\(x^2-5x+5y-y^2\)
\(=(x-y).(x+y)-5.(x-y)\)
\(=(x-y).(x+y-5)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3.(x^2-2xy+y^2-4z^2)\)
\(=3[\left(x-y\right)^2-\left(2z\right)^2]\)
\(=3.(x-y-2z).(x-y+2z)\)
\(x^2+4x+3\)
\(=(x^2+x)+(3x+3)\)
\(=x.(x+1)+3.(x+1)\)
\(=(x+1).(x+3)\)
\((x^2+1)^2-4x^2\)
\(=(x^2-2x+1).(x^2+2x+1)\)
\(=(x-1)^2.(x+1)^2\)
\(x^2-4x-5\)
\(=(x^2+x)-(5x+5)\)
\(=x.(x+1)-5.(x+1)\)
\(=(x-5).(x+1)\)
31 tháng 12 2019
a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7
b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)
c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)
d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)
= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)
= -2(x2 - 2x + 1 + x2 - 1 + x2 + 2x + 1) + 6x2 - 6
= -2(3x2 + 1) + 6x2 - 6
= -6x2 - 2 + 6x2 - 6
= -8
e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2
= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2
= (2x + 7 - 2x + 8)2
= 152 = 225
KT
26 tháng 4 2018
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
KA
8 tháng 11 2021
A=(x−1)2+8≥8Amin=8⇔x=1B=(x+3)2−12≥−12Bmin=−12⇔x=−3C=x2−4x+3+9=(x−2)2+8≥8Cmin=8⇔x=2E=−(x+2)2+11≤11Emax=11⇔x=−2F=9−4x2≤9Fmax=9⇔x=0
HT
8 tháng 11 2021
A=x2-2x+9
Ta có: A=x^2-2x+9
=> A=(x^2-2x+1)+8
=>A=(x-1)^2+8
vì (x-1)^2 > 0 với mọi x
=> (x-1)^2+8> 8 với mọi x
Dấu "=" xáy ra khi:
(x-1)^2=0=>x-1=0=>x=0+1=>x=1
Vậy Amin = 8 khi x=1
B=x^2+6x-3
=>B=-(x^2-6x+3)
=>B=-(x^2-2.3x+3^2)-3
=>B=-(x-3)^2-3
vì -(x-3)^2 < 0 với mọi x
=>-(x-3)^2-3< -3 với mọi x
Dấu '=' xảy ra khi x-3=0=>x=0+3=>x=3
Vậy B(min)=-3 khi x=3
chỗ này hình như là Bmax xem lại đề nhé
D=-x^2-4x+7
=>D=-x^2-2.2x+4+3
=>D=(-x^2-2.2x+4)+3
=>D=(-x-2)^2+3
Vì (-x-2)^2 <0 với mọi x
=>(-x-2)^2+3<3 với mọi x
Dấu "=" xảy ra khi x-2=0=>x=0+2=>x=2
Vậy Dmax=3 khi x=2
E=5-4x^2+4x
=>E=-4x^2+4x+5
=>E=(-2x)^2+2.2x+4+1
=>E=[(-2x)^2+2.2x+4]
=>E=(-2x+2)^2+1
Vì: (-2x+2)^2 < 0 với mọi x
=>(-2x+2)^2+1 < 1 với mọi x
Dấu "=" xảy ra khi 2x+2=0=>2x=-2=>x=-1
Vậy Emax=1 khi x=-1
AT
17 tháng 6 2017
a/ \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)=12x^5-6x^3+x^2\)
b/ \(\left(4x^2+8xy-3xy^2\right)\left(-\dfrac{3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c/ \(4x^3\left(2x^2-x+5\right)5x=20x^4\left(2x^2-x+5\right)\)
\(=40x^6-20x^5+100x^4\)
DH
17 tháng 6 2017
a, \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)\)
\(=12x^5-6x^3+x^2\)
b, \(\left(4x^2+8xy-3xy^2\right).\left(\dfrac{-3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c, \(4x^3\left(2x^2-x+5\right)5x\)
\(=\left(8x^5-4x^4+20x^3\right)5x\)
\(=40x^6-20x^5+100x^4=20x^4.\left(2x^2-x+5\right)\)
Chúc bạn học tốt!!! Mình không chắc đâu !
4 tháng 4 2020
a) (x^5 + 4x^3 - 6x^2) : 4x^2
= (x^5 : 4x^2) + (4x^3 : 4x^2) - (6x^2 : 4x^2)
= 1/4x^3 + x + 3/2
b) x(2x^2 - 3) - x^2(5x + 1) + x^2
= 2x^3 - 3x - 5x^3 - x^2 + x^2
= -3x^3 - 3x
c) (x - 2)^2 - (x - 1)(x + 1) - x(1 - x)
= x^2 - 4x + 4 - x^2 + 1 - x + x^2
<=> x^2 - 5x + 5
d) 1/2x^2(6x - 3) - x(x^2 + 1/2) + 1/2(x + 4)
= \(\frac{x^2}{2}\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{x+4}{2}\)
= \(\frac{x^2\left(6x-3\right)}{2}-x\left(x^2+\frac{1}{2}\right)+\frac{x+4}{2}\)
= \(-x\left(x^2+\frac{1}{2}\right)+\frac{x^2\left(6x-3\right)+x+4}{2}\)
= \(\frac{4x^3-3x^2+4}{2}\)
\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\Leftrightarrow x^2+4x+4+4x=4x^2+8x-3x-6\)
\(\Leftrightarrow3x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{129}}{6}\\x=\frac{3-\sqrt{129}}{6}\end{cases}}\)
\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\)
\(x^2+4x+4+4x=4x^2+8x-3x-6\)
\(3x^2-3x-10=0\)
\(\Delta=\left(-3\right)^2-\left(4.3.-10\right)=129\)
\(\sqrt{\Delta}=\sqrt{129}\)
\(x_1=\frac{3+\sqrt{129}}{6}\)
\(x_2=\frac{3-\sqrt{129}}{6}\)