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

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

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

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

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

\(=4y^2\)

6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)

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

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

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

\(=\left(-10\right)^2=100\)

7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)

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

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

\(=\left(-3\right)^2=9\)

8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)

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

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

\(=\left(4x\right)^2=16x^2\)

Bạn nên tách bài ra để đăng. Không nên đăng 1 loạt như thế này.

1: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

=(x^2+x)^2+3(x^2+x)-10

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

2: \(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)

\(=\left(x^2+5ax+5a^2\right)^2\)

3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)

=n(n+1)(n+2)+360 chia hết cho 6

6A

7D

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)

\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)

\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)

\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)

\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)

\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)

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

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

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

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

\(=5^2=25\)

4) \(\left(3x-5\right)^2-2\left(3x-5\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x-5\right)-\left(3x+5\right)\right]^2\)

\(=\left(3x-5-3x-5\right)^2\)

\(=\left(-10\right)^2\)

\(=100\)

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

\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)

\(\Leftrightarrow-8x-31=0\)

\(\Leftrightarrow x=\dfrac{-31}{8}\)

\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)

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

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

\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)

\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)

\(\Leftrightarrow96x=-117\)

\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)

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

\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(x^2+4x+4\right)\)

\(\Leftrightarrow4x^2+12x+9+x^2-1=5x^2+20x+20\)

\(\Leftrightarrow4x^2+x^2-5x^2+12x-20x=20-9+1\)

\(\Leftrightarrow-8x=12\)

\(\Leftrightarrow x=\dfrac{-12}{8}=\dfrac{-3}{2}\)

bằng 13,34590301 ( mình bấm máy tính bạn nhé :) )

2. CMR:

a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)

Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.

b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)

Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)

=> đpcm.

c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)

\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)

=> đpcm.

1.

b. \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)

\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)

\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)

\(\Leftrightarrow6x=24\)

\(\Leftrightarrow x=4\)

Vậy ....

\(a\text{)}.\: \left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow8x-5x^2+16-10x+4x^2-4x-8+2x^2-8=0\\ \Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

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

\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)

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

\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)