\(A=2x^2+8x-20=2\left(x+2\right)^2-28\)

Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+2\right)^2-28\ge-28\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy Amin = - 28 <=> x = - 2

A = 2x² + 8x - 20

A = 2( x² + 4x + 4 ) - 28

A = 2( x + 2 )² - 28

2( x + 2 )² ≥ 0 ∀ x => 2( x + 2 )² - 28 ≥ -28

Đẳng thức xảy ra <=> x + 2 = 0 => x = -2

=> MinA = -28 <=> x = -2

a)\(\frac{x^2+3x+2}{3x+6}=\frac{x^2+2x+x+2}{3\cdot\left(x+2\right)}=\frac{\left(x^2+2x\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}=\frac{x\cdot\left(x+2\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}\)

\(=\frac{\left(x+2\right)\cdot\left(x+1\right)}{3\cdot\left(x+2\right)}=\frac{x+1}{3}\)

b) \(\frac{2x^2+x-1}{6x-3}=\frac{2x^2+2x-x-1}{3\cdot\left(2x-1\right)}=\frac{\left(2x^2+2x\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}\)

\(=\frac{2x\cdot\left(x+1\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{\left(2x-1\right)\cdot\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{x+1}{3}\)

Bài 1: 

a) (x+y)²=9²=81

=> x²+2xy+y²=81

=> x²+2.14+y²=81

=> x²+y²=53

=> x²-2xy+y²=81-2.14=25

=> (x-y)²=25

=> x-y=5 hoặc x-y=-5

b) Câu a đã tính được x²+y²=53

c) Ta có: x³+y³=(x+y)(x²-xy+y²)=9(53-14)=9.39=351

Bài 2: 

Ta có: \(x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1\)

Mà x+y=1

\(\Rightarrow1^2-4.1+1=-2\)

Bài 3: 

Ta có: (x+y)³=x³+3x²y+3xy²+y³ 

= x³+y³+3xy(x+y)

Mà x+y=1 => (x+y)³=x³+y³+3xy=1³=1

Bài 4: 

Ta có: \(\left(x+y\right)^2=4^2=16\)

\(\Rightarrow x^2+2xy+y^2=16\Rightarrow10+2xy=16\)

\(\Rightarrow2xy=6\Rightarrow xy=3\)

Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4.\left(10-3\right)\)

\(=4.7=28\)

Bài 5: 

Ta có: \(x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=1\left(x^2+xy+y^2\right)-3xy=x^2+xy+y^2-3xy\)

\(=x^2-2xy+y^2=\left(x-y\right)^2=1\)

Mấy bài này đầu hè làm hết rồi:))

Bài 1:

a) \(xy=14\Rightarrow x=\frac{14}{y}\)

Thay vào: \(\frac{14}{y}+y=9\)

\(\Leftrightarrow y^2+14-9y=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2\\y=7\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}\)

+ Nếu: \(\hept{\begin{cases}x=7\\y=2\end{cases}}\Rightarrow x-y=5\)

+ Nếu: \(\hept{\begin{cases}x=2\\y=7\end{cases}}\Rightarrow x-y=-5\)

b) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^2=81\)

\(\Leftrightarrow x^2+2xy+y^2=81\)

\(\Rightarrow x^2+y^2=81-2xy=81-2.14=53\)

c) Ta có: \(x+y=9\)

\(\Leftrightarrow\left(x+y\right)^3=9^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=729\)

\(\Leftrightarrow x^3+y^3=729-3xy\left(x+y\right)=729-3.14.9=351\)

B = x² - 4x + 2

B = ( x² - 4x + 4 ) - 2

B = ( x - 2 )² - 2

( x - 2 )² ≥ 0 ∀ x => ( x - 2 )² - 2 ≥ -2

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinB = -2 <=> x = 2

\(B=x^2-4x+2=\left(x-2\right)^2-2\)

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow\left(x-2\right)^2-2\ge-2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy Bmin = - 2 <=> x = 2

Bài làm:

a) \(x^2-2xy+y^2-zx+yz\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(\left(x-y\right)\left(x-y-z\right)\)

a/ \(x^2-2xy+y^2-zx+yz.\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

c/ \(x^2-y^2-2x-2y.\)

\(=x^2-2x+1-y^2-2y-1\)

\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)

\(=\left(x-1\right)^2-\left(y+1\right)^2\)

\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

Bài làm:

a) \(P=x^2-5x=\left(x^2-5x+\frac{25}{4}\right)-\frac{25}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\le-\frac{25}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x=\frac{5}{2}\)

Vậy \(Min_P=-\frac{25}{4}\Leftrightarrow x=\frac{5}{2}\)

a) P = x² - 5x 

         = ( x² - 5x + 25/4 ) - 25/4

         = ( x - 5/2 )² - 25/4

( x - 5/2 )² ≥ 0 ∀ x => ( x - 5/2 )² - 25/4 ≥ -25/4

Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2

=> MinF = -25/4 <=> x = 5/2

b) Q = x² + 2y² + 2xy - 2x - 6y + 2015 

         = ( x² + 2xy + y² - 2x - 2y + 1 ) + ( y² - 4y + 4 ) + 2010

         = [ ( x + y )² - 2( x + y ) + 1² ] + ( y - 2 )² + 2010

         = ( x + y - 1 )² + ( y - 2 )² + 2010

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x,y\\\left(y-2\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+y-1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y-1=0\\y=2\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

=> MinQ = 2010 <=> x = -1 , y = 2

Bài làm:

a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)

b) \(x\left(x-1\right)=x-1\)

\(\Leftrightarrow x^2-x-x+1=0\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

c) \(5x\left(x-1\right)=1-x\)

\(\Leftrightarrow5x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)

d) \(\left(3x-4\right)^2-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(2x-5\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{4}\end{cases}}\)

\(a,x+5x^2=0< =>x\left(5x+1\right)=0\)

\(< =>\orbr{\begin{cases}x=0\\5x+1=0\end{cases}< =>\orbr{\begin{cases}x=0\\5x=-1\end{cases}< =>\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}}\)

\(b,x\left(x-1\right)=x-1< =>x^2-x=x-1\)

\(< =>x^2-x-x+1=0< =>x\left(x-1\right)-\left(x-1\right)=0\)

\(< =>\left(x-1\right)\left(x-1\right)=0< =>x=1\)

\(c,5x\left(x-1\right)=1-x< =>5x^2-5x=1-x\)

\(< =>5x^2-5x+x-1=0< =>5x^2-4x-1=0\)

\(< =>5x^2-5x+x-1=0< =>5x\left(x-1\right)+x-1=0\)

\(< =>\left(5x+1\right)\left(x-1\right)=0< =>\orbr{\begin{cases}5x+1=0\\x-1=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=1\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=1\end{cases}}}\)

\(d,\left(3x-4\right)^2-\left(x+1\right)^2=0\)

\(< =>9x^2-24x+16-x^2-2x-1=0\)

\(< =>8x^2-26x+15=0< =>8\left(x^2-\frac{13}{4}x+\frac{169}{64}\right)-\frac{2082}{64}=0\)

\(< =>\left(x-\frac{13}{8}\right)^2=\frac{2082}{512}=\frac{2082}{16\sqrt{2}}\)

\(< =>\orbr{\begin{cases}x-\frac{13}{8}=\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x-\frac{13}{8}=-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{13}{8}+\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x=\frac{13}{8}-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)(nghiệm vô tỉ)

a, \(A=\left(2x-3\right)\left(4x+1\right)-4\left(x-1\right)\left(2x-1\right)-2x+5\)\

\(=8x^2+2x-12x-3-8x^2+4x+8x-4-2x+5=-2\)đpcm 

b, \(3x\left(2x-5y\right)+\left(3x+y\right)\left(-2x\right)-\frac{1}{2}\left(2-26xy\right)\)

\(=6x^2-15xy-6x^2-2xy-1+13xy\)

\(=-4xy-1\)Có phụ thuộc 

A = ( 2x - 3 )( 4x + 1 ) - 4( x - 1 )( 2x - 1 ) - 2x + 5

A = 8x² - 10x - 3 - 4( 2x² - 3x + 1 ) - 2x + 5

A = 8x² - 10x - 3 - 8x² + 12x - 4 - 2x + 5

A = ( 8x² - 8x² ) + ( -10x + 12x - 2x ) + ( 5 - 4 - 3 )

A = -2

Vậy A không phụ thuộc vào biến ( đpcm )

B = 3x( 2x - 5y ) + ( 3x - y )( -2x ) - 1/2( 2 - 26xy )

B = 6x² - 15xy - 6x² + 2xy - 1 + 13xy

B = ( 6x² - 6x² ) + ( -15xy + 2xy + 13xy ) - 1

B = -1

Vậy B không phụ thuộc vào biến ( đpcm )

đề khó hiểu quá

2x2 là 2 nhân 2 hay 2 xờ 2

2x²  

Đề bị lỗi 

a) Ta có:

\(A\left(x\right)=x^3-30x^2-31x+1\)

\(A\left(x\right)=x^3-31x^2+x^2-31x+1\)

\(A\left(x\right)=\left(x^3-31x^2\right)+\left(x^2-31x\right)+1\)

\(A\left(x\right)=x^2.\left(x-31\right)+x.\left(x-31\right)+1\)

\(A\left(x\right)=\left(x-31\right).\left(x^2+x\right)+1\)

+ Thay \(x=31\) vào biểu thức \(A\left(x\right)\) ta được:

\(A\left(x\right)=\left(31-31\right).\left(31^2+31\right)+1\)

\(A\left(x\right)=0.992+1\)

\(A\left(x\right)=0+1\)

\(A\left(x\right)=1.\)

Vậy giá trị của biểu thức \(A\left(x\right)\) là \(1\) tại \(x=31.\)