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\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+8abc\)

\(=a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)+8abc\)

\(=ab^2-2abc+ac^2+bc^2-2abc+ba^2+ca^2-2abc+cb^2+8abc\)

\(=ab^2+ac^2+bc^2+ba^2+ca^2+cb^2+2abc\)

\(=\left(ac^2+bc^2\right)+\left(ab^2+ba^2\right)+\left(ca^2+cb^2+2abc\right)\)

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

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

\(=\left(a+b\right)\left[c^2+ab+c\left(a+b\right)\right]=\left(a+b\right)\left(c^2+ab+ca+bc\right)\)

\(=\left(a+b\right)\left[\left(c^2+ca\right)+\left(ab+bc\right)\right]=\left(a+b\right)\left[c\left(c+a\right)+b\left(a+c\right)\right]\)

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

Bài làm

a) x² - 3x + 2 = 0

<=> x² - x - 2x + 2 = 0

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

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

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

Vậy tập nghiệm của phương trình S={2;1}

b) -x² + 5x - 6 = 0

<=> -x² + 2x + 3x - 6 = 0

<=> -x( x - 2 ) + 3( x - 2 ) = 0

<=> ( 3 - x )( x - 2 ) = 0

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

Vậy tập nghiệm của phương trình S={3;2}

c) 4x² - 12x + 5 = 0

<=> 4x² - 10x - 2x + 5 = 0

<=> 2x( 2x - 1 ) - 5( 2x - 1 ) = 0

<=> ( 2x - 5 )( 2x - 1 ) = 0

<=> \(\orbr{\begin{cases}2x-5=0\\2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)

Vậy tập nghiệm phương trình S={\(\frac{5}{2};\frac{1}{2}\)}

d) 2x² + 5x + 3 = 0

<=> 2x² + 2x + 3x + 3 = 0

<=> 2x( x + 1 ) + 3( x + 1 ) = 0

<=> ( 2x + 3 )( x + 1 ) = 0

<=> \(\orbr{\begin{cases}2x+3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-1\end{cases}}}\)

Vậy tập nghiệm phương trình S = { \(-\frac{3}{2};-1\)}

# Học tốt #

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

\(\Rightarrow x\left(x2+6x+9\right)-3x=x3+12x+6x2+8+1\)

\(\Rightarrow x3+6x2+9x-3x=x3+12x+6x2+9\)

\(\Rightarrow x3+6x2+9x-3x-x3-12x-6x2-9=0\)

\(\Rightarrow-6x-9=0\)

\(\Rightarrow-6x=9\)

\(\Rightarrow x=\frac{-9}{6}=\frac{-3}{2}\)

Vậy pt nhận x = -3 / 2 làm nghiệm duy nất

\(x.\left(x+2\right)=x.\left(x+3\right)\)

\(\Leftrightarrow x+2=x+3\)

\(\Leftrightarrow x-x=3-2\)

\(\Leftrightarrow0x=1\)( Vô nghiệm )

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

\(\Leftrightarrow3x+12-x^2-4x=0\)

\(\Leftrightarrow-x^2-x+12=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

Vậy x = 3 ; x = -4

3 (x + 4) − x ^2 − 4x = 0

⇔ 3x + 12 − x^ 2 − 4x = 0 

⇔ −x^ 2 − x + 12 = 0

⇔ x^ 2 + x − 12 = 0 

⇔ x^ 2 + 4x − 3x − 12 = 0

⇔ x (x + 4) − 3 (x + 4) = 0

⇔ (x − 3)(x + 4) = 0

⇔     x − 3 = 0

         x + 4 = 0

⇔     x = 3 

        x = −4

\(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\)

\(\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\)

\(\ge\frac{\left(a+b+\frac{4}{a+b}\right)^2}{2}\)

\(=\frac{25}{2}\) 

tại a=b=¹/₂

thêm ít cách

Cách 1:

Áp dụng BĐT bunhiacopxki ta được:

\(\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]\left(1^2+1^2\right)\ge\left[\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{a}\right)\right]^2\)

\(\Leftrightarrow\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge\left(1+\frac{1}{a}+\frac{1}{b}\right)^2\)(1)

Ta có:\(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\)( tự CM nha )

ÁP dụng BĐT AM-GM ta có:

\(\sqrt{ab}\le\frac{a+b}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge4\)(2)

Thay (2) vào (1) ta được: 

\(\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge25\)

\(\Rightarrow\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{25}{2}\left(đpcm\right)\)

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

Cách 2: 

Đặt \(P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\)

Ta có: \(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}\)

\(=a^2+\frac{2a}{b}+\frac{1}{16b^2}+\frac{15}{16b^2}+b^2+\frac{2b}{a}+\frac{1}{16a^2}+\frac{15}{16a^2}\)

\(=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\left(\frac{2a}{b}+\frac{2b}{a}\right)+\left(\frac{15}{16b^2}+\frac{15}{16a^2}\right)\)

ÁP dụng BĐT AM-GM ta có:

\(a^2+\frac{1}{16a^2}\ge2\sqrt{a^2.\frac{1}{16a^2}}\ge\frac{1}{2}\)(3)

\(b^2+\frac{1}{16b^2}\ge2\sqrt{b^2.\frac{1}{16b^2}}\ge\frac{1}{2}\)(4)

\(\frac{2a}{b}+\frac{2b}{a}\ge2\sqrt{\frac{2a}{b}.\frac{2b}{a}}\ge4\)(5)

\(\frac{15}{16a^2}+\frac{15}{16b^2}\ge2\sqrt{\frac{15.15}{16.16a^2b^2}}=\frac{15}{8ab}\)(1) 

ÁP dụng BĐT AM-GM ta có:

\(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)(2)

Thay (2) vào (1) ta được:

\(\frac{15}{16a^2}+\frac{15}{16b^2}\ge\frac{15}{2}\)(6)

Cộng (3)+(4)+(5)+(6) ta được: 

\(P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}+4=\frac{25}{2}\)

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

Cách 3:Làm tắt thui ạ

Đặt \(P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\)

\(\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}\ge2ab+\frac{2}{ab}+4\)

\(P\ge2\left(ab+\frac{1}{ab}\right)+4\)

\(P\ge2\left(ab+\frac{1}{16ab}+\frac{15}{16ab}\right)+4\)

giống cách 2 rồi làm nốt