a)   \(x^2+2x+1=\left(x+1\right)^2\)

b)   \(9x^2+y^2+6xy=\left(3x+y\right)^2\)

c)   \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

d)   \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

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

f) mk chỉnh lại đề nha:

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

g)  \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

h)  \(x^2-10xy+25y^2=\left(x-5y\right)^2\)

cảm ơn bn nha!

a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)

b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)

c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)

d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)

e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)

f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)

i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)

A),(-2)⁵:(-2)³=(-2)²=4

B) (-y)⁷ :(-y)³=y⁴

a)\(\left(x^4+8x^2+16\right):\left(x^2+4\right)\)

\(=\left(x^2+4\right)^2:\left(x^2+4\right)\)

\(=x^2+4\)

b)\(\left(25-x^2\right):\left(x+5\right)\)

=\(\left(x^2-5^2\right):\left(x+5\right)\)

\(=\left(x-5\right)\left(x+5\right):\left(x+5\right)\)

\(=x-5\)

c)\(\left(x^3+1\right):\left(x^2-x+1\right)\)

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

\(=x+1\)

a) \(\left(x^4+8x^2+16\right):\left(x^2+4\right)\)\(=\left(x^2+4\right)^2:\left(x^2+4\right)\)\(=x^2+4\)

b) \(\left(25-x^2\right):\left(x+5\right)=\left(x-5\right).\left(x+5\right):\left(x+5\right)\)\(=x-5\)

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

Học tốt

đề bài là : dùng hằng đẳng thức để khai triển và thu gọn các biểu thức

ok bạn

\(\left[{}\begin{matrix}A\left(x\right)=x^4-3x^3+ax+b=x^2\left(x^2-3x+4\right)+\left[\left(a-4\right)x+b\right]=B\left(x\right)+f\left(x\right)\left(a\right)\\A\left(x\right)=x^4-3x^3+ax+b=x^2\left(x^2-3x+2\right)+\left[\left(a-2\right)x+b\right]=C\left(x\right)+g\left(x\right)\left(b\right)\end{matrix}\right.\)

a) \(A\left(x\right)⋮B\left(x\right)\Rightarrow f\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}a=4\\b=0\end{matrix}\right.\)

b)\(A\left(x\right)⋮C\left(x\right)\Rightarrow g\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)

Nhân ra thôi chứ sao?

thì bạn nhân đi !

\(\left(a-b\right)^2-\left(b-a\right)\)

\(=\left(a-b\right)^2+\left(a-b\right)\)

\(=\left(a-b\right)\left(a-b+1\right)\)

\(5\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)\)

\(=\left(a+b\right)\left[5\left(a+b\right)-\left(a-b\right)\right]\)

\(=\left(a+b\right)\left[5a+5b-a+b\right]\)

\(=\left(a+b\right)\left[4a+6b\right]\)

Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)

\(< =>A>B\)

a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)

\(\Rightarrow A< B\)

b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

...

\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)

\(\Rightarrow A>B\)

c,d tương tự

Tìm x,biết:

a/ $x+5$$x^2=0$

$\iff x\; (\; 1\; +\; 5x\; )\; =\; 0$

\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0

1) x = 0

2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)

Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)

b/$x+1={\left(x+1\right)}^2$

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

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

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

\(\Leftrightarrow\) x+1 = 0 hoặc x = 0

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

Vậy: S=\(\left\{-1;0\right\}\)

c/ $x^3+x=0$

\(\Leftrightarrow\) x(x² + 1) = 0

\(\Leftrightarrow\) x = 0 hoặc x² + 1 = 0

Ta có : x² + 1 \(\ge\) 0 vs mọi x

Vậy: S = \(\left\{0\right\}\)

d/$5x\left(x-2\right)-\left(2-x\right)=$

\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0

\(\Leftrightarrow\) (x - 2)(5x+1) = 0

\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0

\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)

Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)

g/ $x\left(x-4\right)+{\left(x-4\right)}^2=0$

$\iff \; (x\; -\; 4)(\; x+x-4)\; =\; 0$

$\backslash (\backslash Leftrightarrow\backslash )\; x\; -\; 4\; =\; 0\; hoặc\; 2x-4=0$

$\backslash (\backslash Leftrightarrow\backslash )$ x = 4 hoặc x = 2

Vậy: S= \(\left\{2;4\right\}\)

h/ $x^2-3x=0$

$\iff x\; (x-3)\; =\; 0$

\(\Leftrightarrow\) x = 0 hoặc x = 3

Vậy: S = \(\left\{0;3\right\}\)

Vậy: S= \(\left\{0;3\right\}\)
i/$4$