Bài 2

A/  \(x^2-2xy+y^2-4x+4y-5\)

\(=\left(x^2-2xy+y^2\right)-\left(4x-4y\right)-5\)

\(=\left(x-y\right)^2-4\left(x-y\right)-5\)

\(=\left(x-y\right)\left(x-y-4\right)-5\)

b/ trên máy tính đâu có đặt cột dọc được :v chịu khó tính nháp là ra xD

Bài 3

1/a \(\left(x^2-4x\right)^2+2\left(x-2\right)^2=4^3.\)

\(\left(x^2-4x\right)^2+2\left(x^2-4x+4\right)=64\)

Cho \(x^2-4x\) là S

\(\Rightarrow S^2+2\left(S+4\right)=64\)

\(\Rightarrow S^2+2S+8=64\)

\(\Rightarrow S^2+2S=64-8\)

\(\Rightarrow S^2+2S=56\)

Tính ko ra:v đề có sai ko?

2/  \(2x^2+3y^2+4x=19\)

\(\Rightarrow2x^2+4x=19-3y^2\)

\(\Rightarrow2x^2+4x=21-2-3y^2\)

\(\Rightarrow2x^2+4x+2=21-3y^2\)

\(\Rightarrow2\left(x^2+2x+1\right)=21-3y^2\)

\(\Rightarrow2\left(x+1\right)^2=21-3y^2\)

\(\Rightarrow2\left(x+1\right)^2=3\left(7-y^2\right)\)

Từ đây xét tiếp để ra kq :v

1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²

a)(x+y)²-(x-y)²

=(x+y-x+y)(x+y+x-y)

=2y.2x=4xy

b)(3x+1)²-(x+1)²

=(3x+1-x-1)(3x+1+x+1)

=2x.(4x+2)

=4x(2x+1)

c) x³+y³+z³-3xyz

= (x+y)³- 3xy(x+y) +z³-3xyz

=(x+y+z)( x²+2xy+y²-xz-yz+z²)-3xy(x+y+z)

=(x+y+z)(x²+y²+z²-xy-xz-yz)

Phân tích đa thức sau thành nhân tử :

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

b) \(x^3+y^3+z^3-3xyz\)

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Bài 1:

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

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

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

\(=x^2\)

Bài 2:

đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)

Xét BT trái ta có:

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+5}\)

\(=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}\)

GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến

=> đpcm

Bài 1.

( x - y + z ) + ( z - y )² + ( x - y + z )( 2y - 2z )

= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )²

= [ ( x - y + z ) - ( z - y ) ]² 

= ( x - y + z - z + y )²

= x²

Bài 2. ĐKXĐ tự ghi nhé :))

\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)

\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)

\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)

\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)

\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)

\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)

\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)

=> đpcm

a) x³ + (a+b+c)x²+ (ab+ac+bc)x +abc

= x³ +ax²+bx²+cx²+abx+acx+bcx+abc

=x³+cx²+abx+abc+ax²+acx+bx²+bcx

=x² (x+c) + ab (x+c) +ax (x+c) +bx (x+c)

= (x+c) (x²+ab+ax+bx)

= (x+c) { x(x+b)+a(x+b)}

=(x+c) (x+b) (x+a)

Bài 2:

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

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

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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