pt <=> 4x^2=52x+4y^2+52y=0

<=> (2x-13)^2+(2y+13)^2=2.13^2

Đến đây bạn chặn nó là SCP <=2.13^2 là xong

Ta có:

\(x^3+3x=y^3+3y\)

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

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+3\right)=0\)

\(x^2+xy+y^2+3=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+3>0\)

\(\Leftrightarrow x-y=0\)

\(\Leftrightarrow x=y\)

\(\left(2\right)\Leftrightarrow2x^2=20\)

\(\Leftrightarrow x^2=10\)

\(\Leftrightarrow x=y=\pm\sqrt{10}\)

Vậy ............

đèn giao thông

Đèn giao thông

-16x^2 -8x +1 nhé

\(Đkxđ:\hept{\begin{cases}2x-1>0\\4x-3>0\\x>0\end{cases}\Leftrightarrow x>\frac{3}{4}}\)

Phương trình tương đương với: 

\(\left(\frac{x}{\sqrt{2x-1}}-1\right)+\left(\frac{x}{\sqrt[4]{4x-3}}-1\right)=0\)

\(\Leftrightarrow\frac{x-\sqrt{2x-1}}{\sqrt{2x-1}}+\frac{2-\sqrt[4]{4x-3}}{\sqrt[4]{4x-3}}=0\)

\(\Leftrightarrow\frac{x^2-2x+1}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{x^2-\sqrt{4x-3}}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{x^4-4x+3}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{\left(x-1\right)^2\left(x^2+2x+3\right)}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[\frac{1}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{\left(x+1\right)^2+2}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}\right]=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy .............................