(x+2)(x+3) -(x-2)(x+5)=6

=>x²+5x+6-x²-3x+10=6

=>(x²-x²)+(5x-3x)+(10+6)=6

=>2x+16=6

=>2x=-10

=>x=-5

(x + 2) (x + 3) - (x - 2) (x + 5) = 6

<=> x² + 5x +6 - x² - 3x + 10 = 6

<=> 2x + 16 = 6

<=> 2x = -10

<=> x = -5

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

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

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

=>\(\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

Vậy ...

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

\(\Rightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\)

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

áp dụng các hằng đẳng thức thôi mà :)

a)\(x^2-2x+1=25\)

=>\(\left(x-1\right)^2=25\)

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

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

=>\(3\left[\left(x-1\right)^2-x\left(x-5\right)\right]=1\)

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

=>\(3\left(3x+1\right)=1\)

=>\(3x+1=\frac{1}{3}\)

=>\(3x=\frac{-2}{3}\)

=>\(x=\frac{-2}{9}\)

c)\(\left(5-2x\right)^2-16=0\)

=>\(\left(5-2x\right)^2-4^2=0\)

=>\(\left(5-2x-4\right)\left(5-2x+4\right)=0\)

=>\(\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)

ít thôi bạn ạ

Bạn làm được câu nào thì làm.

a chia 7 dư 3 => a có dạng 7k+3

=>a²=(7k+3)²=49k²+42k+9=49k²+42k+7+2=7(7k²+6k+1)+2

Vậy a² chia 7 dư 2 (đpcm)

\(F=a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(1-1\right)\)(vì a-b=1)

\(F=a^2\left(a+1\right)-b^2\left(b-1\right)+ab\)

\(F=a^3+a^2-b^3+b^2+ab\)

\(F=\left(a^3-b^3\right)+a^2+b^2+ab\)

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

\(F=\left(a^2+ab+b^2\right)+\left(a^2+ab+b^2\right)\)(vì a-b=1)

\(F=2\left(a^2+ab+b^2\right)\)

\(F=2\left(a^2-2ab+b^2+3ab\right)\)

\(F=2\left(\left(a-b\right)^2+3ab\right)\)

\(F=2\left(1+3ab\right)\)

\(F=2+6ab\)

ta có x+y+z=0 

=> \(\left(x+y+z\right)^2=0\)

\(< =>x^2+y^2+z^2+2xy+2xz+2yx=0\)

\(< =>x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(< =>x^2+y^2+z^2+2.0=0\)(vì xy+xz+yz=0)

\(< =>x^2+y^2+z^2=0\)

\(< =>\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}< =>x=y=z=0}\)

thay x=y=z=0 vào 

\(K=\left(x-1\right)^{2014}+y^{2015}+\left(z+1\right)^{2016}\)

\(K=\left(0-1\right)^{2014}+0^{2015}+\left(0+1\right)^{2016}\)

\(K=1+0+1=2\)

\(\)

thanks nhìu

ta có 

\(\left(a+b+c\right)^3=0\)(vì a++c=0)

\(< =>a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)(1)

vì a+b+c= 0 > a+b = -c; a+c = -b ; b+c = -a thay vào (1) ta được

\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(< =>a^3+b^3+c^3+3\left(-c\right)\left(-a\right)\left(-b\right)=0\)

\(< =>a^3+b^3+c^3-3abc=0\)

\(< =>a^3+b^3+c^3=3abc\)