\(\sqrt{2x^2+x+9}+\sqrt{2x^2-x+1}=x+4\)

\(\Leftrightarrow\sqrt{2x^2+x+9}-\left(\frac{1}{2}x+3\right)+\sqrt{2x^2-x+1}-\left(\frac{1}{2}x+1\right)=0\)

\(\Leftrightarrow\frac{2x^2+x+9-\left(\frac{1}{2}x+3\right)^2}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{2x^2-x+1-\left(\frac{1}{2}x+1\right)^2}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}=0\)

\(\Leftrightarrow\frac{\frac{1}{4}x\left(7x-8\right)}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{\frac{1}{4}x\left(7x-8\right)}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}=0\)

\(\Leftrightarrow\frac{1}{4}x\left(7x-8\right)\left(\frac{1}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{1}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}\right)=0\)

Dễ thấy: \(\frac{1}{\sqrt{2x^2+x+9}+\frac{1}{2}x+3}+\frac{1}{\sqrt{2x^2-x+1}+\frac{1}{2}x+1}>0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\7x-8=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{8}{7}\end{cases}}\)

\(\sqrt[3]{x}+\sqrt{5-x}=3\)

Đk:\(0\le x\le5\)

\(pt\Leftrightarrow\sqrt[3]{x}-1+\sqrt{5-x}-2=0\)

\(\Leftrightarrow\frac{x-1}{\sqrt[3]{x}+1}+\frac{5-x-4}{\sqrt{5-x}+2}=0\)

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

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

\(\Rightarrow x=1\). Pt trong ngoặc thì chịu nhưng nghiệm là \(8\sqrt{5}-16\)

à :v sửa tất cả dấu = thành =< nhé nhầm :33

mũ 4 2 vế lên rút gọn là thấy ngay vô nghijem :v

Cần gì phải nâng bậc 4 ^

Đặt biến
\(\left(x+1,x\right)=\left(a,b\right) \\ a+b=2\sqrt[4]{a^4+b^4}\ge a+b \\ \)
Dấu = => \(x=x+1=>Vo.nghiem\)

\(\left(x+3\right)\left(2x-1\right)-\left(x-4\right)\left(2x+1\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-\left(2x^2+x-8x-4\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-2x^2-x+8x+4=10\)

\(\Leftrightarrow12x+1=10\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\frac{9}{12}\)

\(\Rightarrow x=\frac{3}{4}\)

Vậy   \(x=\frac{3}{4}\)

a,\(-\sqrt{10x^2\cdot y\left(3-\sqrt{2}\right)^2}=-\left|x\right|\) \(\cdot\left(3-\sqrt{2}\right)\cdot\sqrt{10y}\)

xet th \(x\ge0\) ta co \(-x\cdot\left(3-\sqrt{2}\right)\sqrt{10y}\)

xet th \(x< 0\) ta có \(x\left(3-\sqrt{2}\right)\sqrt{10y}\)

b,\(\sqrt{3\left(x^2-2xy+y^2\right)}=\) \(\sqrt{3\cdot\left(x-y\right)^2}=\left|x-y\right|\sqrt{3}\)