PTĐTTNT: x5+x4+1
PTĐTTNT: x5+x4+1
Cho x+y=2;x^2+y^2=b;x^3+y^3=c
Chứng minh a^3-3ab+2c=0
Sửa đề: Cho \(x+y=a;x^2+y^2=b;x^3+y^3=c\)
Chứng minh: \(a^3-2ab+2c=0\)
Giải:
Ta có:
\(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=x^3+y^3+3xy\left(x+y\right)-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=3\left(x^3+y^3\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)=3\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)\)
\(=3\left(x+y\right)\left(x^2-xy+y^2+xy-x^2-y^2\right)=3\left(x+y\right).0\)
\(=0\) (đpcm)
Bài 1 : Tìm GTNN
a) Q = \(\dfrac{3}{2}\)x2 + x + 1
b) R = x2 + 2y2 + 2xy - 2y
Bài 2 : Tìm GTLN
a) Q = 2x -2 - 3x2
b) 2 - x2 - y2 - 2(x + y)
c) 7 - x2 - y2 - 2(x + y)
Bài 1:
a ) \(Q=\dfrac{3}{2}x^2+x+1=\dfrac{3}{2}\left(x^2+\dfrac{2}{3}x+\dfrac{2}{3}\right)=\dfrac{3}{2}\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{5}{9}\right)=\dfrac{3}{2}\left[\left(x+\dfrac{1}{3}\right)^2+\dfrac{5}{9}\right]=\dfrac{3}{2}\left(x+\dfrac{1}{3}\right)^2+\dfrac{5}{6}\ge\dfrac{5}{6}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy Min Q là : \(\dfrac{5}{6}\Leftrightarrow x=-\dfrac{1}{3}\)
b ) \(R=x^2+2y^2+2xy-2y=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy Min R là : \(-1\Leftrightarrow x=-1;y=1\)
Bài 2 :
a ) \(Q=2x-2-3x^2\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{5}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{5}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\le-\dfrac{5}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy Max Q là : \(-\dfrac{5}{3}\Leftrightarrow x=\dfrac{1}{3}\)
b ) \(2-x^2-y^2-2\left(x+y\right)\)
\(=2-x^2-y^2-2x-2y\)
\(=-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)+4\)
\(=-\left(x+1\right)^2-\left(y+1\right)^2+4\le4\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-1\)
Vậy Max của b/t trên là : \(4\Leftrightarrow x=-1\)
c ) \(7-x^2-y^2-2\left(x+y\right)\)
\(=7-x^2-y^2-2x-2y\)
\(=-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)+9\)
\(=-\left(x+1\right)^2-\left(y+1\right)^2+9\le9\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-1\)
Vậy Max của b/t trên là : \(9\Leftrightarrow x=y=-1\)
cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) . Tính A=\(\dfrac{x+y}{z}+\dfrac{y+z}{z}+\dfrac{z+x}{y}\)
\(A=\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\) (đã sửa đề)
\(A+3=\dfrac{x+y+z}{z}+\dfrac{x+y+z}{x}+\dfrac{x+y+z}{y}\)
\(A+3=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\)
\(A=-3\)
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Cho: ab+a+b = 3; bc+b+c=8; ca+c+a=15
Tính: S= a+b2 +c2
Cho a^2+b^2= ba + ba. Chứng minh a=b
Very easy :
\(a^2+b^2=ab+ab\)
\(\Leftrightarrow a^2+b^2-2ab=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
\(\Leftrightarrow a=b\left(đpcm\right)\)
Tìm GTLN
a) \(C=-x^2-2x+8\)
b) \(D=-x^2-8+5\)
â, C=-x2-2x+8
=-(x2+2x-8)
=-(x2+2x+1-9)
=-[(x+1)2-9]
=-(x+1)2+9\(\le\)9
=> GTLN C=0( Khi và chỉ khi (x+1)2=0
<=> x=-1)
b, Sửa đề:
D=-x2-8x+5
=-(x2+8x-5)
=-(x2+2.x.4+42-21)
=-[(x+4)2-21]
=-(x+4)2+21\(\le\)21
=> GTLN của D=21( Khi và chỉ khi (x+4)2=0
<=> x=-4)
Đúng thì tick nha,
Tìm GTNN
a) \(A=4x^2+4x+11\)
b) \(B=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
a, A=4x2+4x+11
=(2x)2+2x.2.1+12+10
=(2x+1)2+10\(\ge\)10
=> GTNN của A=10( Khi và chỉ khi (2x+1)2=0
<=> x=-1/2)
b, B=(x2+5x-6)(x2+5x+6)
=(x2+5x)2-62
=(x2+5x)2-36\(\ge\)-36
=> GTNN của B=-36( Khi và chỉ khi (x2+5x)2=0
<=> x={-5;0})
Đúng tick nha,
a, A =4x2 + 4x + 11
A=[(2x)2 + 2.2x + 12 ] + 10
A=(2x + 1)2 +10
có (2x+1)2 ≥ 0 ⇒ (2x+1)2 +10 ≥ 10
do đó GTNN của A là 10 khi x = \(\dfrac{-1}{2}\)
b, B = (x2 + 5x - 6)(x2 + 5x + 6)
B= (x2 + 5x)2 - 62
B= (x2 +5x)2 - 36
có (x2 + 5x) ≥ 0 ⇒(x2 + 5x) - 36 ≥ -36
do đó GTNN của B là -36 khi x ϵ{-5 ; 0}