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\(C=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu '=' xảy ra khi x=1/2
\(D=25x^2-10xy+y^2+2y^2+4y+2-1\)
\(=\left(5x-y\right)^2+2\left(y+1\right)^2-1>=-1\)
Dấu '=' xảy ra khi y=-1 và x=1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)
\(=x^2-3x+6x-12\)
\(=x^2+3x-12\)
b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
c: \(\left(-2xy+3\right)\left(xy+1\right)\)
\(=-2x^2y^2-2xy+3xy+3\)
\(=-2x^2y^2+xy+3\)
d: \(x\left(xy-1\right)\left(xy+1\right)\)
\(=x\left(x^2y^2-1\right)\)
\(=x^3y^2-x\)
Bài 2:
a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
\(=27\cdot\dfrac{1}{27}+8=9\)
b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)
\(=125x^3-8y^3\)
\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=6x^2+4y^2+6xy+\left(xy+\dfrac{4x}{y}\right)+\left(3xy+\dfrac{3y}{x}\right)+2022\)
\(M\ge3x^2+y^2+3\left(x+y\right)^2+2\sqrt{\dfrac{4x^2y}{y}}+2\sqrt{\dfrac{9xy^2}{x}}+2022\)
\(M\ge3\left(x^2+1\right)+\left(y^2+4\right)+3\left(x+y\right)^2+4x+6y+2015\)
\(M\ge6x+4y+3\left(x+y\right)^2+4x+6y+2015\)
\(M\ge3\left(x+y\right)^2+10\left(x+y\right)+2015\ge3.3^2+10.3+2015=2072\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = x2 - 10x + 12
= ( x2 - 10x + 25 ) - 13
= ( x - 5 )2 - 13
( x - 5 )2 ≥ 0 ∀ x => ( x - 5 )2 - 13 ≥ -13
Đẳng thức xảy ra <=> x - 5 = 0 => x = 5
=> MinA = -13 <=> x = 5
B = 6y2 + 4y - 1
= 6( y2 + 2/3y + 1/9 ) - 5/3
= 6( y + 1/3 )2 - 5/3
6( y + 1/3 )2 ≥ 0 ∀ x => 6( y + 1/3 )2 - 5/3 ≥ -5/3
Đẳng thức xảy ra <=> y + 1/3 = 0 => y = -1/3
=> MinB = -5/3 <=> y = -1/3
C = x2 + y2 - 2x - 6y - 1
= ( x2 - 2x + 1 ) + ( y2 - 6y + 9 ) - 11
= ( x - 1 )2 + ( y - 3 )2 - 11
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y-3\right)^2-11\ge-11\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
=> MinC = -11 <=> x = 1 ; y = 3
D = 2x2 + 3y2 - x - 3y + 5
= 2( x2 - 1/2x + 1/16 ) + 3( y2 - y + 1/4 ) + 33/8
= 2( x - 1/4 )2 + 3( y - 1/2 )2 + 33/8
\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x-\frac{1}{4}\right)^2+3\left(y-\frac{1}{2}\right)^2+\frac{33}{8}\ge\frac{33}{8}\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{4}=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{1}{2}\end{cases}}\)
=> MinD = 33/8 <=> x = 1/4 ; y = 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
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1/
( a + b )3 + ( a - b )3 - 6ab2 < đã sửa >
= a3 + 3a2b + 3ab2 + b3 + a3 - 3a2b + 3ab2 - b3 - 6ab2
= 2a3
2/
A = x2 + y2 - 2x - 4y + 6 = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 1 = ( x - 1 )2 + ( y - 2 )2 + 1 ≥ 1 ∀ x, y
Dấu "=" xảy ra khi x = 1 ; y = 2
=> MinA = 1 <=> x = 1 ; y = 2
B = 2x2 + 8x + 10 = 2( x2 + 4x + 4 ) + 2 = 2( x + 2 )2 + 2 ≥ 2 ∀ x
Dấu "=" xảy ra khi x = -2
=> MinB = 2 <=> x = -2
C = 25x2 + 3y2 - 10x + 11 = ( 25x2 - 10x + 1 ) + 3y2 + 10 = ( 5x - 1 )2 + 3y2 + 10 ≥ 10 ∀ x, y
Dấu "=" xảy ra khi x = 1/5 ; y = 0
=> MinC = 10 <=> x = 1/5 ; y = 0
D = ( x - 3 )2 + ( x - 11 )2
Đặt t = x - 7
D = ( t + 4 )2 + ( t - 4 )2
= t2 + 8t + 16 + t2 - 8t + 16
= t2 + 32 ≥ 32 ∀ t
Dấu "=" xảy ra khi t = 0
=> x - 7 = 0 => x = 7
=> MinD = 32 <=> x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
b) \(=x^2-\left(3y\right)^2=\left(x-3y\right)\left(x+3y\right)\)
c) \(=\left(2x-1\right)^2-\left(2y\right)^2=\left(2x-1-2y\right)\left(2x-1+2y\right)\)
d) \(=x^2-10xy+\left(5y\right)^2=\left(x-5y\right)^2\)
e) \(=\left(3x\right)^2-6x+1=\left(3x-1\right)^2\)
f) \(=\left(5x\right)^2+20x+4=\left(5x+2\right)^2\)
\(C=x^2-x+1=x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Nên GTNN của C là \(\frac{3}{4}\) đặt được khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(D=25x^2+3y^2-10xy+4y+1\)
\(=\left(5x\right)^2-2.5x.y+y^2+2y^2+4y+2-1\)
\(=\left(5x-y\right)^2+2\left(y+1\right)^2-1\ge-1\)
Nên GTNN của D là - 1 đạt được khi \(\hept{\begin{cases}y+1=0\\5x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-1\\y=5x\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-1\\x=-\frac{1}{5}\end{cases}}\)