Ta có: \(x+\frac{1}{9}=\frac{8}{x}-1\)

    \(\Leftrightarrow x-\frac{8}{x}=-1-\frac{1}{9}\)

    \(\Leftrightarrow\frac{x^2-8}{x}=-\frac{10}{9}\)

    \(\Leftrightarrow9.\left(x^2-8\right)=-10x\)

    \(\Leftrightarrow9x^2+10x-72=0\)

    \(\Leftrightarrow\left(9x^2+10x+\frac{25}{9}\right)-\frac{673}{9}=0\)

    \(\Leftrightarrow\left(3x+\frac{5}{3}\right)^2=\frac{673}{9}\)

    \(\Leftrightarrow3x+\frac{5}{3}=\pm\frac{\sqrt{673}}{3}\)

+ \(3x+\frac{5}{3}=\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{\sqrt{673}-5}{3}\)

                                                \(\Leftrightarrow\)\(x=\frac{\sqrt{673}-5}{9}\)

+ \(3x+\frac{5}{3}=-\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{-\sqrt{673}-5}{3}\)

                                                \(\Leftrightarrow\)\(x=\frac{-\sqrt{673}-5}{9}\)

Vậy \(x\in\left\{\frac{\sqrt{673}-5}{9};\frac{-\sqrt{673}-5}{9}\right\}\)

Ta có : \(\frac{x+1}{9}=\frac{8}{x-1}\left(x\ne1\right)\)

\(\Rightarrow\left(x+1\right).\left(x-1\right)=8.9\)

\(\Rightarrow x^2-1=72\Rightarrow x^2=73\Rightarrow x=\pm\sqrt{73}\)

Áp dụng liên tiếp BĐT quen thuộc \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\) ta được :

\(\left(a^2+b^2\right)+\left(c^2+d^2\right)\) \(\ge\frac{\left(a+b\right)^2}{2}+\frac{\left(c+d\right)^2}{2}\)

\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2}{2}\ge\frac{\frac{\left(a+b+c+d\right)^2}{2}}{2}=\frac{\left(a+b+c+d\right)^2}{4}=\frac{2^2}{4}=1\)

Do đó : \(a^2+b^2+c^2+d^2\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d=\frac{1}{2}\)

Theo Svacxo ta có : \(LHS\ge\frac{\left(a+b+c+d\right)^2}{4}=\frac{2^2}{4}=1\left(đpcm\right)\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=\frac{1}{2}\)

Có :

\(A=n^3-7n\)

\(=\left(n^3-n\right)-6n\)

\(=n.\left(n^2-1\right)-6n\)

\(=\left(n+1\right)n\left(n-1\right)-6n⋮6\)

\(A=n^3-7n\)

\(=n^3-n-6n\)

\(=\left(n^3-n\right)-6n\)

\(=n\left(n^2-1\right)-6n\)

\(=\left(n+1\right)n\left(n-1\right)-6n⋮6\)

\(\Rightarrow A⋮6\left(dpcm\right)\)

\(3^{2n+1}+5.2^{3n+1}\)

Với \(n=1\)thì \(3^5+5.2^4=243+80=323⋮19\)

Gải sử \(3^{2k+1}+5.2^{3k+1}⋮19\)

Xét \(3^{3k+5}+5.2^{3k+4}=3^{3k+2}.3^3+5.2^{3k+1}.2^3\)

\(=27\left(3^{3k+2}+5.2^{3k+1}\right)-19.3^{2k+1}⋮19\)

Đặt \(S_n=3^{2n+1}+40n-67\)

Xét \(n=1\Rightarrow S_n=0⋮64\)

Giả sử n đúng với \(n=k\left(k\inℤ^+\right)\)tức là ta có :

\(S_k=3^{2k+1}+40k-67⋮64\). Ta cần chứng minh n đúng với \(n=k+1\).

Tức cần chứng minh \(S_{k+1}=2^{2\left(k+1\right)+1}+40\left(k+1\right)-67⋮64\)

Thật vậy ta có : \(S_{k+1}=2^{2\left(k+1\right)+1}+40\left(k+1\right)-67\)

\(=9\cdot2^{2k+1}+40k-27\)

\(=9\cdot\left(2^{2k+1}+40k-67\right)-320k+576\)

\(=9\cdot S_k-320k+576⋮64\)

Vậy n đúng với \(n=k+1\)

Do đó \(S_n=3^{2n+1}+40n-67⋮64\forall n\inℤ^+\)

Với \(n=1\)thì \(3^3+40-67=0⋮64\)

Giả sử \(3^{2k+1}+40k-67⋮64\)

Xét \(3^{2k+3}+40\left(k+1\right)-67\)

\(=9\left(3^{2k+1}+40k-67\right)+64\left(9-5k\right)⋮64\)

\(\)

\(S_n=\frac{1.2.3.4...n\left(n+1\right)\left(n+2\right)...2n}{1.2.3.4...n}\)

\(=\frac{1.3...\left(2n-1\right).2.4...\left(2n-2\right)2n}{1.2.3.4...n}\)

\(=\frac{1.3...\left(2n-1\right).2^n.1.2...n}{1.2...n}\)

\(=2^n.1.3...\left(2n-1\right)⋮2n\)

\(A=2x^2+2xy+y^2-2x+2y+2\)

\(=\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+x^2-4x+4-3\)

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

\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\forall x,y\)

Dấu"="xảy ra khi \(\orbr{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}y=-3\\x=2\end{cases}}}\)

Vậy.....

A = 2x² + 2xy + y² - 2x + 2y + 2

= ( x² + 2xy + y² + 2x + 2y + 1 ) + ( x² - 4x + 4 ) - 3

= [ ( x + y )² + 2( x + y ) + 1² ] + ( x - 2 )² - 3

= ( x + y + 1 )² + ( x - 2 )² - 3 ≥ -3 ∀ x, y

Dấu "=" xảy ra <=> x = 2 ; y = -3

=> MinA = -3 <=> x = 2 ; y = -3

B thì nhờ các cao nhân khác ._. Em tịt rồi

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy \(x=3\)hoặc \(x=-5\)

\(x^2-3x+5\left(x-3\right)=0\) 

\(x^2-3x+5x-15=0\) 

\(x^2+2x-15=0\)  

\(x^2-3x+5x-15=0\) 

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

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

\(\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\) 

\(\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

a) x² - xy - 20y²

= x² + 4xy - 5xy - 20y²

= x( x + 4y ) - 5y( x + 4y )

= ( x + 4y )( x - 5y )

b) x³ - x²y - 3xy² + 2y³

= x³ + x²y - 2x²y - xy² - 2xy² + 2y³

= ( x³ + x²y - xy² ) - ( 2x²y + 2xy² - 2y³ )

= x( x² + xy - y² ) - 2y( x² + xy - y² )

= ( x² + xy - y² )( x - 2y )