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 \(x=31.\)

giúp mk vs

Bạn viết rõ đề ra được không . MÌnh không hiểu đề cho lắm

Đặt \(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}\right)^2-1^2\)

     \(2A=5^{32}-1\)

\(\Rightarrow A=\frac{5^{32}-1}{2}.\)

N = (-y² + 4)(2y³ + 6y - 1) + 2(y⁵ - 4y³ + 2)-  y²(-6y + 1)

N = -y²(2y³ + 6y - 1) + 4(2y³ + 6y - 1) + 2y⁵ - 8y³ - 4 + 6y³ - y²

N = -2y⁵ - 6y³ + y² + 8y³ + 24y - 4 + 2y⁵ - 8y³ - 4 + 6y³ - y²

N = (-2y⁵ + 2y⁵) + (-6y³ + 8y³ - 8y³ + 6y³) + (y² - y²) + 24y + (-4 - 4)

N = 24y - 8

Thay y = -3,5 vào biểu thức N ta có :

N = 24.(-3,5) - 8 = -84 - 8 = -92

Ta có : \(x^2+x+4=x^2+x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+x=-4\end{cases}}\)

+) x² + x = - 4

<=> ( x + 1/2 )² = - 4 + 1/4 = -15/4

Mà ( x + 1/2 )² lớn hơn hoặc bằng 0 với mọi x

=> x² + x + 4 = 0 ktm

Vậy pt = 0 <=> x = 1