Những hằng đẳng thức đáng nhớ

Ta có: a+b+c=0

nên \(\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Leftrightarrow2ab+2ac+2bc=-1\)

\(\Leftrightarrow ab+ac+bc=\dfrac{-1}{2}\)

\(\Leftrightarrow\left(ab+ac+bc\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{1}{4}\)

Ta có: \(a^2+b^2+c^2=1\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1\)

\(\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\)

\(\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)

Vậy: \(a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}\)

(x-3)² - (2x-1)(2x+1) = 10

<=> x² - 6x + 9 - 4x² + 1 = 10

<=> -3x² - 6x = 0

<=> x² + 2x = 0

<=> x(x + 2) = 0

<=> \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

vậy ....

\(4(x-3)\)\(^2\) \(-(2x-1)(2x+1)=10\)

\(=> 4(x\)\(^2\) \(-6x+9)-4x\)\(^2\)\(-1=10\)

\(=>\)\(4x\)\(^2\)\(-24x+36-4x\)\(^2\) \(+1=10\)

            \(=>-24x+27=10\)

            \(=>-24x=-27\)

             \(=>x=\)\(\dfrac{-27}{-24}\)\(=\) \(\dfrac{9}{8}\)

                             \(Vậy \)\(X=\)\(\dfrac{9}{8}\)

         Nhớ vote cho mik hen

144=\(12^2\)

\(144=12^2=\left(-12\right)^2\)

Ai giúp mik với ạ 😢😭😭😭😭😢😷

\(A=\left(x^2+y^2+36-2xy-12x+12y\right)+5y^2-10y+5+109\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+109\ge109\)

\(A_{min}=109\) khi \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

6x^2-(x-3)(3x+2)-6

=6x^2-6x^2+x-9x-6-6

=-5x-12

     \(\left(2x-1\right)^{^3}\)

=  \(\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)

=  \(8x^3-12x^2+6x-1\)

8x³-12x²+6x-1

\(A=x^2-2xy+2y^2+2x-10y+2033\\ =x^2-2xy+y^2+y^2+2x-8y-2y+1+16+2016\\ =\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(y^2-8y+16\right)+2016\\ =\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-4\right)^2+2016\\ =\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]+\left(y-4\right)^2+2016\\ =\left(x-y+1\right)^2+\left(y-4\right)^2+2016\\ Do\text{ }\left(y-4\right)^2\ge0\forall y\\ \left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ \Rightarrow A=\left(x-y+1\right)^2+\left(y-4\right)^2+2016\ge2016\forall x;y\\ Dấu\text{ }''=''\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}\left(y-4\right)^2=0\\\left(x-y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-4=0\\x-y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x-4+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\\ Vậy\text{ }A_{\left(Min\right)}=2016\text{ }khi\text{ }\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

xem lại đề

Chắc là \(x^3-3x^2-x+3=0\)

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

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

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

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

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

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

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

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

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