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LD
31 tháng 8 2018
a,
\(x^2+4y^2-x+4y+2=\left(x^2-x+\dfrac{1}{4}\right)+4\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+4\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x,y\)
b,
\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0-3\left(-c\right)\left(-a\right)\left(-b\right)=0-3\left(-abc\right)=3abc\left(dpcm\right)\)
KK
2
13 tháng 9 2017
a+b+c+d=0
=>a+b=-(c+d)
=>(a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=>a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=>a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) vi a+b=-(c+d)
=> a^3+b^3+c^3+d^3=3(c+d)(ab+cd)
Xem lai gium mk nha!!
26 tháng 9 2017
Ta có : a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd)
10 tháng 1 2018
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
CT
1
22 tháng 10 2017
\(\text{a) }4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[\left(2x\right)^2-5^2\right]-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\\ \Leftrightarrow-2\left(2x-5\right)=0\\ \Leftrightarrow2x-5=0\\ \Leftrightarrow2x=5\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }x=\dfrac{5}{2}\\ \)
\(\text{b) }2x^3+3x^2+2x+3=0\\ \Leftrightarrow\left(2x^3+3x^2\right)+\left(2x+3\right)=0\\ \Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(KTM\right)\\2x=-3\left(TM\right)\end{matrix}\right.\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }x=-\dfrac{3}{2}\)
HA
Tìm x biết:
a, ( 2x - 5)2 = ( x - 2)2
b, ( x + 1)( 2 - x) - ( 3x + 5)( x + 2) = - 4x2 + 1
Help me!!!
4
22 tháng 10 2017
\(a,\left(2x-5\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left(2x-5\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\3x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\)
\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12+4x^2-1=0\)
\(\Rightarrow-10x-13=0\)
\(\Rightarrow x=-\dfrac{13}{10}\)
AH
Akai Haruma
Giáo viên
11 tháng 9 2020
Lời giải:
a)
$x^3+y^3+2x^2-2xy+2y^2=(x^3+y^3)+2(x^2-xy+y^2)$
$=(x+y)(x^2-xy+y^2)+2(x^2-xy+y^2)=(x^2-xy+y^2)(x+y+2)$
b)
$a^4+ab^3-a^3b-b^4=(a^4-a^3b)+(ab^3-b^4)$
$=a^3(a-b)+b^3(a-b)=(a-b)(a^3+b^3)=(a-b)(a+b)(a^2-ab+b^2)$
c)
\(a^3-b^3+3a^2+3ab+3b^2=(a^3-b^3)+3(a^2+ab+b^2)\)
\(=(a-b)(a^2+ab+b^2)+3(a^2+ab+b^2)=(a^2+ab+b^2)(a-b+3)\)
d)
\(x^4+x^3y-xy^3-y^4=x^3(x+y)-y^3(x+y)=(x+y)(x^3-y^3)=(x+y)(x-y)(x^2+xy+y^2)\)
\(x^3+2x-3=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)
Ta có: \(x^2+x+3=\left[x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+3-\left(\dfrac{1}{2}\right)^2\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\) (1)
Mà \(\left(x-1\right)\left(x^2+x+3\right)=0\) từ (1) \(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy x = 1
\(x^3-x+3x-3=0\)
\(\Leftrightarrow x\left(x^2-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+3\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vl\right)\end{matrix}\right.\)
vậy \(S=\left\{1\right\}\)