\(x^3-7x+6=\left(x^3-x^2\right)+\left(x^2-x\right)-\left(6x-6\right)\)

\(=x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-6\right)\)

\(=\left(x^2+x-6\right)\left(x-1\right)\)

\(=\left(x^2+3x-2x-6\right)\left(x-1\right)\)

\(=\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\left(x-1\right)\)

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

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

x^2 - 2xy + 6y^2 - 12x + 2y +45 
= x^2 - 2x(y+6) + (y+6)^2 - (y+6)^2 + 6y^2 +2y + 45 
= (x - y - 6)^2 - y^2 - 12y - 36 + 6y^2 + 2y + 45 
= (x - y - 6)^2 + 5y^2 - 10y + 9 
= (x - y - 6)^2 + 5.(y^2 - 2y +1) + 4 
= (x - y - 6)^2 + 5.(y-1)^2 + 4 
=>> MIN = 4 khi (x;y) = {(7;1)}

Sửa đề:\(A=x^2-2xy+6y^2-12x-2y+45\)

Ta có:\(A=x^2-2xy+6y^2-12x-2y+45\)

\(=\left(x^2-2xy+y^2\right)-\left(12x+12y\right)+36+\left(5y^2-10y+5\right)+4\)

\(=\left(x-y\right)^2-12\left(x-y\right)-6^2+5\left(y^2-2y+1\right)+2^2\)

\(=\left[\left(x-y\right)^2-12\left(x-y\right)-6^2\right]+5\left(y-1\right)^2+4\)

\(=\left(x-y+6\right)^2+5\left(y-1\right)^2+4\ge4\forall x,y\)

Dấu "=" xảy ra khi \(\orbr{\begin{cases}x-y+6=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\y=1\end{cases}}}\)

Vậy....

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)ĐK : \(x\ne1\)

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

\(\Rightarrow3x^2+x-4=4x-4\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là S = { 0 } 

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)ĐKXĐ:x khác 1

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

\(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)

\(\Leftrightarrow3x^2+x-4-4x+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)

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

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

\(x^2-4x+3=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy phương trình có tập nghiệm : \(S=\left\{1;3\right\}\)

\(B=-2x^2-x+\frac{25}{8}=-\left(2x^2+x+\frac{1}{8}\right)+\frac{13}{4}=-\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{13}{4}\le\frac{13}{4}\)

Dấu = xảy ra khi:

\(\sqrt{2}x+\frac{1}{2\sqrt{2}}=0\)

\(\Leftrightarrow x=-\frac{1}{4}\)

ko biết

1963 chia 7 dư 3

$\Rightarrow 19631964 chia\; 7\; dư\; 31964$

Mà 3¹⁹⁶⁴  = 9⁹⁸²

9 chia 7 dư 2$\Rightarrow$9⁹⁸² chia 7 dư 2⁹⁸²

Mà 2⁹⁸²=2.8³²⁷

8 chia 7 dư 1 $\Rightarrow$ 8³²⁷ chia cho 7 dư 1³²⁷=1

$\Rightarrow$ 2.8³²⁷ chia cho 7 dư 2

$\Rightarrow$ 1963¹⁹⁶⁴ chia cho 7 dư 2

sao mà khó thế?

A B C H K M I

a, Xét tam giác BAC và tam giác AHC ta có : 

^BAC = ^AHC = 90⁰

^C _ chung 

Vậy tam giác BAC ~ tam giác AHC ( g.g )

b, Xét tam giác AHB và tam giác HKA ta có 

^BHA = ^HKA = 90⁰

^BAH = ^AHK ( so le trong )

Vậy tam giác AHB = tam giác HKA ( g.g )

\(\Rightarrow\frac{AH}{HK}=\frac{AB}{AH}\)( tỉ số tương ứng ) \(\Rightarrow AH^2=AB.HK\)