hộ mình với

\(2x-3=\frac{x+1}{2}\)

\(\Rightarrow2\left(2x-3\right)=x+1\)

\(\Rightarrow4x-6=x+1\)

\(\Rightarrow3x=7\)

\(\Rightarrow x=\frac{7}{3}\)

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

\(\Leftrightarrow3x^2+3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

1.theo bất đẳng thức côsi ta có

\(a+b\ge2\sqrt{ab}\\ b+c\ge2\sqrt{ab}\\ c+a\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab.bc.ca}\)

                                       \(\ge8\sqrt{a^2b^2c^2}\\ \ge8abc\)

2.\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^4b^2\)

\(\dfrac{a}{a^4+b^2}\le\dfrac{a}{2a^2b}=\dfrac{1}{2ab}\)

tương tự:\(\dfrac{b}{b^4+a^2}\le\dfrac{1}{2ab}\)

\(\rightarrow\dfrac{a}{a^4+b^2}+\dfrac{b}{b^4+a^2}\le\dfrac{1}{ab}\)

dấu = xảy ra khi \(a^4=b^2\\ b^4=a^2\)\(\rightarrow a^2=b^2=1\)

ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(\Leftrightarrow x\sqrt{3x-2}-x^2+\left(x+1\right)\sqrt{5x-1}-\left(x+1\right)^2+x^2+\left(x+1\right)^2-8x+3=0\)

\(\Leftrightarrow x\left(\sqrt{3x-2}-x\right)+\left(x+1\right)\left(\sqrt{5x-1}-x-1\right)+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\dfrac{-x\left(x^2-3x+2\right)}{\sqrt{3x-2}+x}+\dfrac{-\left(x+1\right)\left(x^2-3x+2\right)}{\sqrt{5x-1}+x+1}+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)+\left(2-\dfrac{x}{\sqrt{3x-2}+x}-\dfrac{x+1}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{\sqrt{3x-2}}{\sqrt{3x-2}+x}+\dfrac{\sqrt{5x-1}}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\) (ngoặc đằng sau luôn dương)

\(\Leftrightarrow...\)

a.

\(\Leftrightarrow\left(sinx+cosx\right)\left(1+sinx.cosx\right)=1\)

Đặt \(sinx+cosx=t\) \(\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(t\left(1+\dfrac{t^2-1}{2}\right)=1\)

\(\Leftrightarrow t^3+t-2=0\)

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

\(\Leftrightarrow t=1\)

\(\Rightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}=sin\left(\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow...\)

b.

Đặt \(sinx-cosx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Phương trình trở thành:

\(t^3=1+\dfrac{1-t^2}{2}\)

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}=sin\left(\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow...\)

đề bài bảo j vậy

\(2x^2+xy-10x-5y\)

\(=\left(2x^2+xy\right)-\left(10x+5y\right)\)

\(=x\left(2x+y\right)-5\left(2x+y\right)\)

\(=\left(2x+y\right)\left(x-5\right)\)

\(x^2-y^2+9-6x\)

\(=\left(x^2-6x+9\right)-y^2\)

\(=\left(x-3\right)^2-y^2\)

\(=\left(x-3-y\right)\left(x-3+y\right)\)

\(8,=\left(2x-3\right)\left(2x+3\right)\\ 9,=\left(1-5a^2\right)\left(1+5a^2\right)\)

8) \(-9+4x^2=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

9) \(1-25a^4=1-\left(5a^2\right)^2=\left(1-5a^2\right)\left(1+5a^2\right)\)

ko có ảnh bn ơi

Đề bài là gì zạ?