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1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
A) x2+4y22+z22-4x-6z+15>0 <=> (x2-2×2×x+22)+4y2+(z2-2×3×z+32) +(15 -22-32) >0
<=>(x-2)2+4y22+(z-3)2
B) giải
(2X)2+ 2×2X×1 +1 >=0 với mọi X ( (2x+1)2 )
=> (2x+1)2+2 >0
A=x2-2x+2
A=(x2-2x+1)+1
A=(x-1)2+1
(x-1)2\(\ge\)0 với mọi x
=> (x-1)2+1 >0 hay A>0
Vậy A luôn dương với mọi x,y,z
B=x2+y2+z2+4x-2y-4z+10
B=(x2+4x+4)+(y2-2y+1)+(z2-4z+4)+1
B=(x+2)2+(y-1)2+(z-2)2+1
(x+2)2\(\ge\)0 với mọi x
(y-1)2\(\ge\)0 với mọi y
(z-2)2\(\ge\)0 với mọi z
=>(x+2)2+(y-1)2+(z-2)2+1>0 hay B>0
Vậy B luôn dương với mọi x,y,z
C=x2+y2+2x-4y+6
C=(x2+2x+1)+(y2-4y+4)+1
C=(x+1)2+(y-2)2+1
(x+1)2\(\ge\)0 với mọi x
(y-2)2\(\ge\)0 với mọi y
=>(x+1)2+(y-2)2+1>0 hay C>0
Vậy C luôn dương với mọi x,y,z
a/ \(A=x^2-2x+2\\A=x^2-2x+1+1\\ A=\left(x-1\right)^2+1>0 \)
b/ \(B=x^2+y^2+z^2+4x-2y-4z+10\)
\(B=x^2+4x+4+y^2-2y+1+z^2-4z+4+1\)
\(B=\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2+1>0\)
c/ \(C=x^2+y^2+2x-4y+6\)
\(C=x^2+2x+1+y^2-4y+4+1\)
\(C=\left(x+1\right)^2+\left(y-2\right)^2+1>0\)
b)5x^2+9y^2-12xy-6x+9=0
=>4x^2-12xy+9y^2+x^2-6x+9=0
=>(2x-3y)^2+(x-3)^2=0
=>2x-3y=0 và x-3=0
=>y=2 và x=3
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
c: =>(2x+3y-1)^2+(2x-3y)=0
=>2x-3y=0 và 2x+3y=1
=>x=1/4; y=1/6
d: =>2y-3=0 và 2x+3y-1=0
=>y=3/2 và 2x=1-3y=1-9/2=-7/2
=>x=-7/4 và y=3/2
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a, B=x2+4xy+y2+x2-8x+16+2012
B=(x+y) 2+(x-4)2+2012
Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)
b làm tương tự
c, 9x2+6x+1+y2-4y+4+x2-4xz+4z2=0
(3x+1)2+(y-4)2+(x-2z)2=0
Vậy 3x+1=0 => x = -1/3
y-4=0 => y=4
x-2z=0 thế x=-1/3 ta được. -1/3-2z=0 => z = -1/6
Bạn nhớ ghi lại đề minh không ghi đề
a) \(B=2x^2+y^2+2xy-8x+2028\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)
\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)
b)\(C=x^2+5y^2+4xy+2x+2y-7\)
\(=\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+\left(y^2-2y+1\right)-9\)
\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)
\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)
A=x 2−2x+2
=x2-2x+1+1
=(x2-2x+1)+1
=(x-1)2+1
vì (x-1)2\(\ge0\forall x\)
=>(x-1)2+1\(\ge1\)
vậy A luôn dương với mọi x
B=x2+y2+2x−4y+6
=x2+2x+1+y2-4y+4+1
=(x2+2x+1)+(y2-4y+4)+1
=(x+1)2+(y-2)2+1
do (x+1)2\(\ge0\forall x\)
(y-2)2\(\ge0\forall y\)
=>(x+1)2+(y-2)2\(\ge0\)
=>(x+1)2+(y-2)2+1\(\ge1\)
=>B\(\ge1\)
vậy B luôn dương với mọi x;y
C= x2+y2+z2+4x−2y−4z+10
=x2+4x+4+y2-2y+1+z2-4z+4+1
=(x2+4x+4)+(y2-2y+1)+(z2-4z+4)+1
=(x+2)2+(y-1)2+(z-2)2+1
do (x+2)2\(\ge0\forall x\)
(y-1)2\(\ge0\forall y\)
(\(\)z-2)2\(\ge0\forall z\)
=>(x+2)2+(y-1)2+(z-2)2\(\ge0\)
=>(x+2)2+(y-1)2+(z-2)2+1\(\ge1\)
=>C\(\ge1\)
vậy C luôn dương với mọi x;y;z
bài 2: tìm x
a)\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow x^2+y^2-2x+4y+1+4=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy x=1; y=-2
b)\(5x^2+9y^2-12xy-6x+9=0\)
\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(2x-3y\right)^2+\left(x-3\right)^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2.3-3.y=0\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)
Vậy x=2; y=3