(x-1)\(^3\)+(x-2)\(^3\)+(3-2x)\(^3\)=0
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xét tứ giác ABCD, có: góc A + B + C + D=360*
ta có: A/1 = B/2 = C/3 = D/4
xét tc dãy ts = nhau, có:
A+B+C+D/1+2+3+4 = 360/10 = 36
=> A=36
=> B=36.2=72
=> C=36.3=108
=>D=36.4=144
Xét tứ giác ABCD ta có :
^A + ^B + ^C + ^D = 3600(định lí tổng các góc trong một tứ giác)
Mà ^A : ^B : ^C : ^D = 1 : 2 : 3 : 4 => \(\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{D}}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{D}}{4}=\frac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{1+2+3+4}=\frac{360^0}{10}=36^0\)
Từ đó suy ra ^A = 360 . 1 = 360 , ^B = 360 . 2 = 720 , ^C = 360 . 3 = 1080 , ^D = 360 . 4 = 1440
Đến đây tự kết luận
a) 27x3 - 27x2 + 18x - 4 ( sửa thành +18x nhé , chứ để như kia khó phân tích lắm :< )
= 27x3 - 18x2 - 9x2 + 12x + 6x - 4
= ( 27x3 - 18x2 + 12x ) - ( 9x2 - 6x + 4 )
= 3x( 9x2 - 6x + 4 ) - ( 9x2 - 6x + 4 )
= ( 9x2 - 6x + 4 )( 3x - 1 )
b) 2x3 - x2 + 5x + 3
= 2x3 - 2x2 + x2 + 6x - x + 3
= ( 2x3 - 2x2 + 6x ) + ( x2 -x + 3 )
= 2x( x2 - x + 3 ) + ( x2 - x + 3 )
= ( x2 - x + 3 )( 2x + 1 )
c) ( x2 - 3 )2 + 16
= x4 - 6x2 + 9 + 16
= x4 - 6x2 + 25
= x4 + 10x2 + 25 - 16x2
= ( x4 + 10x2 + 25 ) - 16x2
= ( x2 + 5 )2 - ( 4x )2
= ( x2 - 4x + 5 )( x2 + 4x + 5 )
Ta có: \(m^2-2n^2=mn\)
\(\Leftrightarrow m^2-2n^2-mn=0\)
\(\Leftrightarrow m^2-n^2-n^2-mn=0\)
\(\Leftrightarrow\left(m^2-n^2\right)-\left(n^2-mn\right)=0\)
\(\Leftrightarrow\left(m-n\right)\left(m+n\right)-n\left(n-m\right)=0\)
\(\Leftrightarrow\left(m-n\right)\left(m+n\right)+n\left(m-n\right)=0\)
\(\Leftrightarrow\left(m-n\right)\left(m+n+n\right)=0\)
\(\Leftrightarrow\left(m-n\right)\left(m+2n\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-n=0\\m+2n=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=n\\m=-2n\end{cases}}\)
TH1: Nếu \(m=n\)\(\Rightarrow m-n=0\)\(\Rightarrow A=\frac{m-n}{m+n}=0\)
TH2: Nếu \(m=-2n\)\(\Rightarrow A=\frac{-2n-n}{-2n+n}=\frac{-3n}{-n}=3\)
Vậy nếu \(m=n\)thì \(A=0\)
nếu \(m=-2n\)thì \(A=3\)
Câu 1.
P = x2 - 2x + 5
= ( x2 - 2x + 1 ) + 4
= ( x - 1 )2 + 4 ≥ 4 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinP = 4 <=> x = 1
Q = 2x2 - 6x
= 2( x2 - 3x + 9/4 ) - 9/2
= 2( x - 3/2 )2 - 9/2 ≥ -9/2 ∀ x
Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2
=> MinQ = -9/2 <=> x = 3/2
M = x2 + y2 - x + 6y + 10
= ( x2 - x + 1/4 ) + ( y2 + 6y + 9 ) + 3/4
= ( x - 1/2 )2 + ( y + 3 )2 + 3/4 ≥ 3/4 ∀ x
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
=> MinM = 3/4 <=> x = 1/2 ; y = -3
Câu 2.
A = 4x - x2 + 3
= -( x2 - 4x + 4 ) + 7
= -( x - 2 )2 + 7 ≤ 7 ∀ x
Đẳng thức xảy ra <=> x - 2 = 0 => x = 2
=> MaxA = 7 <=> x = 2
B = x - x2
= -( x2 - x + 1/4 ) + 1/4
= -( x - 1/2 )2 + 1/4 ≤ 1/4 ∀ x
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
=> MaxB = 1/4 <=> x = 1/2
N = 2x - 2x2
= -2( x2 - x + 1/4 ) + 1/2
= -2( x - 1/2 )2 + 1/2 ≤ 1/2 ∀ x
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
=> MaxB = 1/2 <=> x = 1/2
Làm gần xong thì lỡ bấm out ra TT
\(P=x^2-2x+5=\left(x-1\right)^2+4\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy minP = 4 <=> x = 1
\(Q=2x^2-6x=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow2\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
Vậy minQ = - 9/2 <=> x = 3/2
\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)
Vì \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
Vậy minM = 3/4 <=> x = 1/2 và y = - 3
212 = ( 20 + 1 )2 = 202 + 2.20.1 + 12 = 400 + 40 + 1 = 441
1992 = ( 200 - 1 )2 = 2002 - 2.200.1 + 12 = 40 000 - 400 + 1 = 39 601
299.301 = ( 300 - 1 )( 300 + 1 ) = 3002 - 12 = 90 000 - 1 = 89 999
312 = ( 30 + 1 )2 = 302 + 2.30.1 + 12 = 900 + 60 + 1 = 961
992 = ( 100 - 1 )2 = 1002 - 2.100.1 + 12 = 10 000 - 200 + 1 = 9801
62.58 = ( 60 + 2 )( 60 - 2 ) = 602 - 22 = 3600 - 4 = 3596
1992 = (200 – 1)2 = 2002 – 2.200 + 1 = 40000 – 400 + 1 = 39601
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=\left(x^2-4y^2\right)^2\)
\(=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{x^2y}{2}+\frac{x^2y}{2}-\frac{y^2}{4}=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
\(P=\left(x^2+2xy\right)^2+2\left(x^2+2xy\right)y^2+y^4\)
\(=x^4+4x^3y+4x^2y^2+2x^2y^2+4xy^3+y^4\)
\(=x^4+y^4+6x^2y^2+4x^3y+4xy^3\)
P = ( x2 + 2xy )2 + 2( x2 + 2xy )y2 + y4
= ( x2 + 2xy )2 + 2( x2 + 2xy )y2 + ( y2 )2
= ( x2 + 2xy + y2 )2
= [ ( x + y )2 ]2
= ( x + y )4
Đặt \(x-1=a;x-2=b;3-2x=c\)
\(\Rightarrow a^3+b^3-\left(a+b\right)^3=0\)
Đến đây thì dễ rồi :))
Cách trâu bò nhất : Phá tung nó ra =))
( x - 1 )3 + ( x - 2 )3 + ( 3 - 2x )3 = 0
<=> x3 - 3x2 + 3x - 1 + x3 - 6x2 + 12x - 8 - 8x3 + 36x2 - 54x + 27 = 0
<=> ( x3 + x3 - 8x3 ) + ( -3x2 - 6x2 + 36x2 ) + ( 3x + 12x - 54x ) + ( -1 - 8 + 27 ) = 0
<=> -6x3 + 27x2 - 39x + 18 = 0
<=> -3( 2x3 - 9x2 + 13x - 6 ) = 0
<=> -3( 2x3 - 3x2 - 6x2 + 9x + 4 - 6 ) = 0
<=> -3[ ( 2x3 - 3x2 ) - ( 6x2 - 9x ) + ( 4x - 6 ) ] = 0
<=> -3[ x2( 2x - 3 ) - 3x( 2x - 3 ) + 2( 2x - 3 ) ] = 0
<=> -3( 2x - 3 )( x2 - 3x + 2 ) = 0
<=> -3( 2x - 3 )( x2 - x - 2x + 2 ) = 0
<=> -3( 2x - 3 )[ x( x - 1 ) - 2( x - 1 ) ] = 0
<=> -3( 2x - 3 )( x - 1 )( x - 2 ) = 0
<=> \(\hept{\begin{cases}2x-3=0\\x-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=1\\x=2\end{cases}}\)( Thay bằng dấu hoặc hộ mình nhé )
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