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\(f\left(x^5+y^5+y\right)=x^3f\left(x^2\right)+y^3f\left(y^2\right)+f\left(y\right)\)

Sửa lại đề câu 2 !!

Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}};b=\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(a^3+b^3=\frac{12+\sqrt{135}+12-\sqrt{135}}{3}=\frac{24}{3}=8\) ; \(ab=\sqrt[3]{\frac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{3\cdot3}}=\sqrt[3]{\frac{144-135}{9}}=1\)

\(x=\frac13\left(1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)

=>\(3x=1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(3x-1=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(\left(3x-1\right)=a+b\)

=>\(\left(3x-1\right)^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=a^3+b^3+3ab\left(3x-1\right)\)

=>(3x-1)^3=8+3*1*(3x-1)=8+3(3x-1)

=>\(27x^3-27x^2+9x-1=9x+5\)

=>\(27x^3-27x^2=6\)

=>\(9x^3-9x^2=2\)

\(M=\left(9x^3-9x^2-3\right)^2\)

\(=\left(2-3\right)^2=\left(-1\right)^2=1\)

\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)

\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)

+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)

\(x^3=\)

\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)

\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)

\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)

\(x^3=4-3.2x\)

\(x^3=4-6x\)

thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)

\(A=3\left(x-3\right)\left(x+7\right)+\left(x+4\right)^2+48\)

\(A=3\left(x^2-4x-21\right)+\left(x^2+8x+16\right)+48\)

\(A=\left(3x^2+x^2\right)-\left(12x-8x\right)-\left(21-16-48\right)\)

\(A=4x^2-4x+43\)

\(A=\left(4x^2-4x+1\right)+42\)

\(A=\left(2x+1\right)^2+42\)

Thay \(x=\frac{1}{2}\) vao A ta duoc:

\(A=\left(2\cdot\frac{1}{2}+1\right)^2+42=46\)

\(A=3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\)

\(=3x^2-13x+63+x^2+8x+16+48\)

\(=4x^2-5x+127\)

\(4\cdot0,25-5\cdot0,5+127=1-1+127=127\)

ĐKXĐ: \(x\ge-2\)

- Với \(-2\le x< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}>1\Rightarrow\sqrt{x^2+1}-x>1\\\sqrt{x+3}\ge1\Rightarrow\sqrt{x+2}+\sqrt{x+3}\ge1\end{matrix}\right.\)

\(\Rightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x+2}+\sqrt{x+3}\right)>1\) pt vô nghiệm

- Với \(x\ge0\)

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

\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+3}=x+\sqrt{x^2+1}\)

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+3}+x-\sqrt{x+2}=0\)

\(\Leftrightarrow\frac{x^2-x-2}{\sqrt{x^2+1}+\sqrt{x+3}}+\frac{x^2-x-2}{x+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{x+\sqrt{x+2}}\right)=0\)

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

Vậy pt có nghiệm duy nhất \(x=2\)

Dạ em cảm ơn Anh ạ

b: =>|x+2|+|2x-1|<x+1(1)

Trường hợp 1: x<-2

(1) sẽ là -x-2-2x+1<x+1

=>-3x-1<x+1

=>-4x<2

hay x>-1/2(loại)

Trường hợp 2: -2<=x<1/2

(1) sẽ là x+2+1-2x<x+1

=>-x+3<x+1

=>-2x<-2

hay x>1(loại)

Trường hợp 3: x>=1/2

(1) sẽ là x+2+2x-1<x+1

=>3x+1<x+1

=>x<0(loại)

Vậy: BPT vô nghiệm

b: =>|x+2|+|2x-1|<x+1(1)

Trường hợp 1: x<-2

(1) sẽ là -x-2-2x+1<x+1

=>-3x-1<x+1

=>-4x<2

hay x>-1/2(loại)

Trường hợp 2: -2<=x<1/2

(1) sẽ là x+2+1-2x<x+1

=>-x+3<x+1

=>-2x<-2

hay x>1(loại)

Trường hợp 3: x>=1/2

(1) sẽ là x+2+2x-1<x+1

=>3x+1<x+1

=>x<0(loại)

Vậy: BPT vô nghiệm

giống Nguyễn Lê Phước Thịnh nhé