\(\cdot\left(x+1\right)^2\ge0\)

\(\Rightarrow x^2+2x+1>0\)

\(\Rightarrow2x^2+4x+2\ge0\)

 \(\Rightarrow\left(3x^2+3x+3\right)-\left(x^2-x+1\right)\ge0\)

\(\Rightarrow3\left(x^2+x+1\right)\ge x^2-x+1\)

\(\Rightarrow\)\(\frac{x^2+x+1}{x^2-x+1}\ge\frac{1}{3}\) (1)

\(\cdot\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2x^2-4x+2\ge0\)

\(\Rightarrow\left(3x^2-3x+3\right)-\left(x^2+x+1\right)\ge0\)

\(\Rightarrow3\left(x^2-x+1\right)\ge x^2+x+1\)

\(\Rightarrow\frac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ(1),(2) => đpcm

\(\text{Áp dụng BĐT Bunhia... cho 2 bộ số (a;b;c) và (x;y;z), ta có: }\)

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\ge\left(ax+by+cz\right)^2\)

\(\text{Dấu = xảy ra }\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\text{(đpcm)}\)

Chả biết có đúng không '-'

Sửa lại đề:\(\left(ax+by+cz\right)\rightarrow\left(ax+by+cz\right)^2\)

Ta có:\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2aybx-2bzcy-2azcx=0\)

\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

Vì\(\left(ay-bx\right)^2\ge0\)

   \(\left(bz-cy\right)^2\ge0\)

    \(\left(az-cx\right)^2\ge0\)

Suy ra:\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2\ge0\)

Mà\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}ay-bx=0\\bz-cy=0\\az-cx=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ay=bx\\bz=cy\\az=cx\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\)\(\left(x,y,z\ne0\right)\)

\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\left(đpcm\right)\)

Vậy...

Linz

(x - 1)(x + 1)(x² + 1)

Áp dụng HĐT số 3 : (A + B)(A - B) = A² - B²

= (x² - 1²)(x² + 1) = (x² - 1)(x² + 1) = (x²)² - 1² = x⁴ - 1

( x - 1 )( x + 1 )( x² + 1 )

= ( x² - 1 )( x² + 1 )

= x⁴ - 1 

2x³ + x - 4x² - 2 = 0

⇔ ( 2x³ - 4x² ) + ( x - 2 ) = 0

⇔ 2x²( x - 2 ) + 1( x - 2 ) = 0

⇔ ( x - 2 )( 2x² + 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2=0\\2x^2+1=0\end{cases}}\)

+) x - 2 = 0 ⇔ x = 2

+) 2x² + 1 = 0

⇔ 2x² = -1 ( vô lí do 2x² ≥ 0 ∀ x )

Vậy phương trình có nghiệm duy nhất là x = 2

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x^2+1=0\left(loai\right)\end{cases}\Leftrightarrow}x=2\)

2x^2 + 3x - ( x^2 + 3x + 2 ) = 6 

2x^2 + 3x - x^2 - 3x - 2 = 6 

x^2 - 2 = 6 

x^2 = 8 

x = +- căn 8 

Ta có : x(2x + 3) - (x + 2)(x + 1) = 6

=> 2x² + 3x - (x² + 3x + 2) = 6

=> 2x² + 3x - x² - 3x - 2 = 6

=> x² = 8

=> \(x=\pm\sqrt{8}\)

a) x(x - 2) + (x - 2) = 0

=> (x + 1)(x - 2) = 0

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

Vậy \(x\in\left\{-1;2\right\}\)

b) \(\frac{2}{3}x\left(x^2-4\right)=0\)

=> x(x² - 4) = 0

=> \(\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=2^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

g) (x + 2)² - x + 4 = 0

=> x² + 4x + 4 - x + 4 = 0

=> x² + 3x + 8 = 0

=> (x² + 3x + 9/4) + 23/4 = 0

=> (x + 3/2)² + 23/4 \(\ge\frac{23}{4}>0\)

=> Phương trình vô nghiệm

h) (x + 2)² = (2x - 1)² 

=> (x + 2)² - (2x - 1)² = 0

=> (x + 2 - 2x + 1)(x + 2 + 2x - 1) = 0

=> (-x + 3)(3x + 1) = 0

=> \(\orbr{\begin{cases}-x+3=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{3}\end{cases}}\)

=> \(x\in\left\{3;-\frac{1}{3}\right\}\)

a) x( x - 2 ) + x - 2 = 0

⇔ x( x - 2 ) + 1( x - 2 ) = 0

⇔ ( x - 2 )( x + 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b) 2/3x( x² - 4 ) = 0

⇔ \(\orbr{\begin{cases}\frac{2}{3}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

g) ( x + 2 )² - x + 4 = 0

⇔ x² + 4x + 4 - x + 4 = 0

⇔ x² + 3x + 8 = 0 (*)

Ta có : x² + 3x + 8 = ( x² + 3x + 9/4 ) + 23/4 = ( x + 3/2 )² + 23/4 ≥ 23/4 > 0 ∀ x

=> (*) không xảy ra 

=> Pt vô nghiệm

h) ( x + 2 )² = ( 2x - 1 )²

⇔ ( x + 2 )² - ( 2x - 1 )² = 0

⇔ [ ( x + 2 ) - ( 2x - 1 ) ][ ( x + 2 ) + ( 2x - 1 ) ] = 0

⇔ ( x + 2 - 2x + 1 )( x + 2 + 2x - 1 ) = 0

⇔ ( 3 - x )( 3x + 1 ) = 0

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

\(a+3\text{ chia hết cho 5 do đó:}a\text{ chia 5 dư 2};\text{ }b+4\text{ chia hết cho 5 nên }b\text{ chia 5 dư 1}\)

\(\text{ do đó:}a^2+b^2\equiv2^2+1^2\equiv5\equiv0\left(\text{mod 5}\right)\text{ ta có điều phải chứng minh}\)

Vì \(a+3⋮5\)\(\Rightarrow\)\(a\)có dạng \(a=5m+2\)( \(m\inℤ\))

    \(b+4⋮5\)\(\Rightarrow\)\(b\)có dạng \(b=5n+4\)( \(n\inℤ\) )

\(a^2+b^2=\left(5m+2\right)^2+\left(5n+1\right)^2\)

\(=25m^2+20m+4+25n^2+10n+1\)

\(=25m^2+20m+25n^2+10n+5⋮5\)( đpcm )

Ta có: \(x^3+y^3+z^2=3xyz+1\)

   \(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=1\)

   \(\Leftrightarrow\left(x+y+z\right)^3-3xy\left(x+y+z\right)-3z\left(x+y\right)\left(x+y+z\right)=1\)

   \(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(zx+zy\right)-3xy\right]=1\)

   \(\Leftrightarrow\left(x+y+z\right)\left[x^2+y^2+z^2+2xy+2yz+2zx-3xy-3yz-3zx\right]=1\)

   \(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=1\)

Đến đây các bạn tự giải nhé ^_^

Bn gì ơi, đây kh pk mk nhờ bn giải hộ, mk nổi hứng đăng câu hỏi lên thôi nên lm hết đi nhá

Sử dụng BĐT Cauchy Schwarz ta dễ có:

\(P=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)

\(=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)

\(\ge\frac{\left(x+y\right)^2}{x+y-2}\)

Ta cần chứng minh: \(\frac{\left(x+y\right)^2}{x+y-2}\ge8\)

\(\Leftrightarrow\left(x+y\right)^2-8\left(x+y\right)+16\ge0\)

\(\Leftrightarrow\left(x+y-4\right)^2\ge0\)( ĐPCM )

Có : \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)

\(=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)

Theo BĐT Cô - si ta có :

\(\frac{x^2}{y-1}+4\left(y-1\right)\ge2\sqrt{\frac{x^2}{y-1}.4\left(y-1\right)}=4x\)

\(\frac{y^2}{x-1}+4\left(x-1\right)\ge4y\)

Do đó ; \(\frac{x^2}{y-1}+\frac{y^2}{x-1}+4.\left(x+y-2\right)\ge4\left(x+y\right)\)

\(\Leftrightarrow\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge8\)

Hay : \(P\ge8\)

Dấu "=" xảy ra khi \(x=y=2\)

Vậy \(P_{min}=8\) khi \(x=y=2\)