\(P=\left(\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-1\right)\)

\(=\left(\dfrac{2}{x-3}-\dfrac{x^2+3}{x^2-9}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x+1}\right)\)

\(=\dfrac{2\left(x+3\right)-x^2-3+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x-1-2x-1}{2x+1}\)

\(=\dfrac{2x+6-x^2-3+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)

\(=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)

\(=\dfrac{-2}{\left(x+3\right)}\times\dfrac{2x+1}{-2}\)

\(=\dfrac{2x+1}{x+3}\)

`a, (2x-3)(x+5)-(x+4)(3-x)=3(x-1)(x+1)`

`<=>2x^2+10x-3x-15-3x+x^2-12+4x=3x^2-3`

`<=>8x=24`

`<=>x=3`

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`b, (4x-3)(x-1)-4x(x-2)=5`

`<=>4x^2-4x-3x+3-4x^2+8x=5`

`<=>x=2`

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`c, (x-2)(x^2+2x+4)-x(x-2)(x+3)=x^2`

`<=>x^3-8-x^3-3x^2+2x^2+6x=x^2`

`<=>2x^2-6x+8=0`

`<=>x^2-3x+4=0`

`<=>x^2-2x .3/2+9/4+7/4=0`

`<=>(x-3/2)^2=-7/4` (Vô lí)

`=>` Ptr vô nghiệm

\(a.\left(2x-3\right)\left(x+5\right)-\left(x+4\right)\left(3-x\right)=3\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x^2-3x+10x-15-\left(3x-x^2+12-4x\right)=3\left(x^2-1\right)\)

\(\Leftrightarrow2x^2+7x-15-\left(-x^2-x+12\right)=3x^2-3\)

\(\Leftrightarrow2x^2+7x-15+x^2+x-12=3x^2-3\)

\(\Leftrightarrow3x^2+8x-27-3x^2+3=0\)

\(\Leftrightarrow8x-24=0\)

\(\Leftrightarrow x=3\)

\(b.\left(4x-3\right)\left(x-1\right)-4x\left(x-2\right)=5\)

\(\Leftrightarrow4x\left(x-1\right)-3\left(x-1\right)-4x.x-4x\left(-2\right)=5\)

\(\Leftrightarrow4x^2-4x-3x+3-4x^2+8x=5\)

\(\Leftrightarrow x=2\)

Gọi 3 số lẻ liên tiếp là n; n+2; n+4, theo đề bài

\(n\left(n+2\right)+n\left(n+4\right)+\left(n+2\right)\left(n+4\right)=3n\left(n+2\right)\)

\(\Leftrightarrow n\left(n+4\right)-n\left(n+2\right)+\left(n+2\right)\left(n+4\right)-n\left(n+2\right)=0\)

\(\Leftrightarrow n\left(n+4-n-2\right)+\left(n+2\right)\left(n+4-n\right)=0\)

\(\Leftrightarrow2n+4n+8=0\Leftrightarrow6n=-8\) (xem lại đề bài)

\(\left(\dfrac{7y+1}{y^2-7y}+\dfrac{7y-1}{y^2+7y}\right)\times\dfrac{y^2-49}{y^2+1}\)

(ĐKXĐ: \(y\ne0;y\ne\pm7\))

\(=\left(\dfrac{7y+1}{y.\left(y-7\right)}+\dfrac{7y-1}{y.\left(y+7\right)}\right)\times\dfrac{\left(y-7\right).\left(y+7\right)}{y^2+1}\)

\(=\dfrac{\left(7y+1\right).\left(y+7\right)+\left(7y-1\right).\left(y-7\right)}{y.\left(y-7\right).\left(y+7\right)}\times\dfrac{\left(y-7\right).\left(y+7\right)}{y^2+1}\)

\(=\dfrac{7y^2+49y+y+7+7y^2-49y-y+7}{y.\left(y-7\right).\left(y+7\right)}\times\dfrac{\left(y-7\right).\left(y+7\right)}{y^2+1}\)

\(=\dfrac{14y^2+14}{y.\left(y-7\right).\left(y+7\right)}\times\dfrac{\left(y-7\right).\left(y+7\right)}{y^2+1}\)

\(=\dfrac{14.\left(y^2+1\right)}{y.\left(y-7\right).\left(y+7\right)}\times\dfrac{\left(y-7\right).\left(y+7\right)}{y^2+1}\)

\(=\dfrac{14.\left(y^2+1\right).\left(y-7\right).\left(y+7\right)}{y.\left(y-7\right).\left(y+7\right).\left(y^2+1\right)}\)

\(=\dfrac{14}{y}\)

Có a + b +c =0

 <=> a + b = -c

=> a²+b²+2ab = c² 

=> a²+b² -c² = -2ab

=>(  a²+b² -c²) = 4a²b²

<=> a⁴ +b⁴ +c⁴+2a²b^2-2b²c²-2a²c²=4a²b²

=> a⁴+b⁴+c⁴= 2a²b²+2b²c²+2c²a²

=> 2(a⁴+b⁴+c⁴)=2a²b²+2b²c²+2c²a²+a⁴+b⁴+c⁴

<=> 2(a⁴+b⁴+c⁴)=(a²+b²+c²)²

<=>a⁴+b⁴+c⁴= \(\dfrac{\left(a^2+b^2+c^2\right)^2}{2}=\dfrac{1^2}{2}=\dfrac{1}{2}\)

Ta có

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

\(\Rightarrow2\left(ab+bc+ac\right)=-1\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)

Ta có

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

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

Ta có

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

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

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\) Thay vào (1)

\(\Rightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1-2x\dfrac{1}{4}=\dfrac{1}{2}\)

Bạn cần hỗ trợ bài nào thì nên ghi chú rõ ra.

Đề như này đúng không bạn

\(A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}\)

 áp dụng bất đẳng thức bunhia dạng phân thức có 

  \(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{3^2}{2\cdot3}=\dfrac{3}{2}\)

Dấu ''='' xay ra khi x=y=z=1

Vậy Amin=3/2 khi x=y=z=1

\(A=\Sigma\dfrac{x^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{x+y+y+z+x+z}=\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{3^2}{2.3}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow x=y=z=1\)

đề thế này à bạn nãy mình đọc chắc nhầm

\(\Sigma\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge\Sigma a-\dfrac{a^2b}{2\sqrt{ab}}=\Sigma a-\dfrac{a\sqrt{ab}}{2}\)

\(\Rightarrow\Sigma\dfrac{a}{ab+1}\ge a+b+c-(\dfrac{a\sqrt{ab}}{2}+\dfrac{b\sqrt{bc}}{2}+\dfrac{c\sqrt{ca}}{2})\)

\(=3-\dfrac{1}{2}\left(\Sigma a\sqrt{ab}\right)\ge3-\dfrac{1}{2}\left(\dfrac{a\left(a+b\right)}{2}+\dfrac{b\left(b+c\right)}{2}+\dfrac{c\left(c+a\right)}{2}\right)=3-\dfrac{1}{4}\left(a^2+b^2+c^2+ab+bc+ca\right)=3-\dfrac{1}{4}\left[\left(a+b+c\right)^2-\left(ab+bc+ca\right)\right]\ge3-\dfrac{1}{4}\left[3^2-\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{1}{4}\left[3^2-\dfrac{3^2}{3}\right]=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(B=\Sigma\dfrac{1}{b+1}\ge\dfrac{9}{a+b+c+1+1+1}=\dfrac{9}{3+3}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=1\)