\(\left[3\left(x-1\right)^2+6\right]\left(3+6\right)\ge\left[3\left(x-1\right)+6\right]^2\)

\(\Leftrightarrow3x^2-6x+9\ge x+5\)

\(\Rightarrow A\ge x^4-8x^2+2024=\left(x^2-4\right)^2+2008\ge2008\)

Dấu "=" xảy ra khi \(x=2\)

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

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

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

\(x^3=\)

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

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

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

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

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

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

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

bạn viết lại đề đi

đề là tìm x ạ

ĐKXĐ: \(x\ge2\)

\(\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=5\)

- TH1:

\(2\le x\le6\)

\(\Rightarrow2-\sqrt{x-2}+3-\sqrt{x-2}=5\)

\(\Leftrightarrow\sqrt{x-2}=0\Rightarrow x=2\)

TH2: \(6\le x\le11\)

\(\Rightarrow\sqrt{x-2}-2+3-\sqrt{x-2}=5\Leftrightarrow1=5\) (vô nghiệm)

TH3: \(x>11\)

\(\Rightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=5\)

\(\Leftrightarrow2\sqrt{x-2}=10\Rightarrow\sqrt{x-2}=5\)

\(\Rightarrow x=27\)

Vậy \(\left[{}\begin{matrix}x=2\\x=27\end{matrix}\right.\)

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

\(pt\text{⇔}\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\text{⇔}x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\\ \text{⇔}17x=17\text{⇔}x=1\)

Vậy nghiệm của phương trình : \(S=\left\{1\right\}\)

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x=17\)

hay x=1

Đề bài yêu cầu gì?

\(a,=\left(x-5\right)\left(x+5\right)\\ b,=\left(x-3\right)^2\\ c,=\left(3x-2\right)\left(3x+2\right)\\ d,=\left(x+1\right)^2\\ e,=\left(x-10\right)\left(x+10\right)\)