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Bài 1 :
1) 4x2 - y2 = ( 2x + y ) ( 2x - y )
2) 9x2 - 4y2 = ( 3x - 2y ) ( 3x + 2y )
3) 4x2 + y2 + 4xy = ( 2x + y )2
Bài 2:
1) 2x2 + 8x = 0
=> 2x ( x + 4 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
2) 3 ( x - 4 ) + x2 - 4x = 0
=> 3 ( x - 4 ) + x ( x - 4 ) = 0
=> ( x - 4 ) ( 3 + x ) = 0
=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
3) 3 ( x - 2 ) = x2 - 2x
=> 3 ( x - 2 ) - x2 + 2x = 0
=> 3 ( x - 2 ) - x ( x - 2 ) = 0
=> ( x - 2 ) ( 3 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
4) x ( x - 2 ) - 6 ( 2 - x ) = 0
=> x ( x - 2 ) + 6 ( x - 2 ) = 0
=> ( x - 2 ) ( x + 6 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
5) 2x ( x + 5 ) = x2 + 5x
=> 2x ( x + 5 ) - x2 - 5x = 0
=> 2x ( x + 5 ) - x ( x + 5 ) = 0
=> ( x + 5 ) ( 2x - x ) = 0
=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
6 ) ( x - 2 )2 - x ( x + 3 ) = 9
=> x2 - 4x + 4 - x2 - 3x = 9
=> - 7x + 4 = 9
=> - 7x = 5
=> x = \(-\frac{5}{7}\)
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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...............<Giải thích như câu đầu>..............
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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..............<Giải thích như câu đầu>......................
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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...............<Giải thích như câu đầu>..............
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
\(D=x^2+y^2+xy-12x+12y+100\)
\(\Rightarrow2D=x^2+x^2+y^2+y^2+2xy-24x+24y+200\)
\(\Rightarrow2D=\left(x^2+2xy+y^2\right)+\left(x^2-2.x.12+12^2\right)+\left(y^2+2.x.12+12^2\right)-88\)
\(\Rightarrow2D=\left(x+y\right)^2+\left(x-12\right)^2+\left(y+12\right)^2-88\)
\(\Rightarrow2D\ge-88\Leftrightarrow D\ge-44\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x-12=0\\x+12=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=12\\y=-12\end{cases}}\)
Vậy : GTNN của \(D=-44\) tại \(x-12,y=-12\).
Bài làm
\(D=x^2+y^2+xy-12x+12y+100\)
Nhân thêm 4 vào đẳng thức trên, ta được
\(4D=4x^2+4y^2+4xy-48x+48y+400\)
\(=\left(4x^2+2.2xy+y^2\right)-24\left(2x+y\right)+3y^2-24y+400\)
\(=\left(2x+y\right)^2-2\left(2x+y\right).12+12^2+3y^2-24y+256\)
\(=\left(2x+y-12-\right)^2+3\left(y-4\right)^2+208\ge208\)
\(\Rightarrow D\ge208:4\)
\(\Rightarrow D=52\)
Dấu " = " xảy ra <=> x = 4; y = 4
Vậy giá trị của biển thứ D = 52 khi x = 4; y = 4
~ Khôg hiểu chỗ nào hỏi mik ~
# Họk tốt #
a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
\(A=x^2+9x+25\)
\(=x^2+2x\frac{9}{2}+\frac{81}{4}+\frac{19}{4}\)
\(=\left(x+\frac{9}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
Dấu"="xảy ra khi \(\left(x+\frac{9}{2}\right)^2=0\Rightarrow x=\frac{-9}{2}\)
Vậy \(Min_A=\frac{19}{4}\Leftrightarrow x=\frac{-9}{2}\)
b,\(B=4x^2-8x+\frac{21}{2}\)
\(=4\left(x^2-2x+1\right)+\frac{13}{2}\)
\(=4\left(x-1\right)^2+\frac{13}{2}\ge\frac{13}{2}\forall x\)
Dấu"="xảy ra khi \(4\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Min_B=\frac{13}{2}\Leftrightarrow x=1\)
c,\(C=-x^2+2x+\frac{5}{2}\)
\(=-\left(x^2-2x-\frac{5}{2}\right)\)
\(=-\left(x^2-2x+1\right)+\frac{7}{2}\)
\(=-\left(x-1\right)^2+\frac{7}{2}\le\frac{7}{2}\forall x\)
Dấu"="xảy ra khi \(-\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy\(Max_C=\frac{7}{2}\Leftrightarrow x=1\)
Bài 1.
A = x2 + 9x + 25
= ( x2 + 9x + 81/4 ) + 19/4
= ( x + 9/2 )2 + 19/4 ≥ 19/4 ∀ x
Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2
=> MinA = 19/4 <=> x = -9/2
B = 4x2 - 8x + 21/2
= 4( x2 - 2x + 1 ) + 13/2
= 4( x - 1 )2 + 13/2 ≥ 13/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinB = 13/2 <=> x = 1
C = -x2 + 2x + 5/2
= -( x2 - 2x + 1 ) + 7/2
= -( x - 1 )2 + 7/2 ≤ 7/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MaxC = 7/2 <=> x = 1
D = -9x2 - 12x + 27/2
= -9( x2 + 4/3x + 4/9 ) + 35/2
= -9( x + 2/3 )2 + 35/2 ≤ 35/2 ∀ x
Đẳng thức xảy ra <=> x + 2/3 = 0 => x = -2/3
=> MaxD = 35/2 <=> x = -2/3
Bài 2.
a) 4x2 + 9y2 + 12x + 12y + 13 = 0
<=> ( 4x2 + 12x + 9 ) + ( 9y2 + 12y + 4 ) = 0
<=> ( 2x + 3 )2 + ( 3y + 2 )2 = 0 (*)
\(\hept{\begin{cases}\left(2x+3\right)^2\ge0\forall x\\\left(3y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(2x+3\right)^2+\left(3y+2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}2x+3=0\\3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{2}{3}\end{cases}}\)
=> x = -3/2 ; y = -2/3
b) 16x2 + 4y2 - 8x + 12y + 10 = 0
<=> ( 16x2 - 8x + 1 ) + ( 4y2 + 12y + 9 ) = 0
<=> ( 4x - 1 )2 + ( 2y + 3 )2 = 0 (*)
\(\hept{\begin{cases}\left(4x-1\right)^2\ge0\forall x\\\left(2y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(4x-1\right)^2+\left(2y+3\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}4x-1=0\\2y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{3}{2}\end{cases}}\)
=> x = 1/4 ; y = -3/2
câu 1
a, 5x - x 2 + 2xy - 5y
= 5x - x 2 + xy + xy - 5y
= ( 5x - 5y ) - ( x2 - xy ) + xy
= 5 ( x-y ) - x(x-y ) + xy
= (5-x) ( x-y) + xy
mik làm dc mỗi câu a !
1)
2x^2 + 12 + 2x.(4-x) = 0
2x^2 + 12 + 8x - 2x^2 = 0
12 + 8x = 0
8x = -12
x = -4/3
2)
A= 8x3 - 12x2y +6xy2 - y3
A = (2x)3 - 3.(2x)2.y + 3.2.x.y2 - y3
A = (2x-y)3
...
tính giá trị mk nhường bn làm đó!
\(B=\left(x^2-8x\right)\left(x^2-8x+24\right)\)
Đặt \(x^2-8x+12=t\) ta có:
\(B=\left(t-12\right)\left(t+12\right)=t^2-144\ge-144\)
Dấu "=" xảy ra khi \(t^2=0\Leftrightarrow t=0\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\Leftrightarrow x=2;x=6\)
\(C=5x^2+9y^2-6xy-12x+13\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-12x+9\right)+4\)
\(=\left(x-3y\right)^2+\left(2x-3\right)^2+4\ge4\)
Dấu "=" xảy ra tại \(x=\frac{3}{2};y=\frac{1}{2}\)