Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)

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

\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)

\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)

\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)

Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình

Ta có:

7.5^2=4.5^2+6^2

→BC2=AB2+AC2

→ΔABC vuông tại A

Ta có BD là phân giác góc B→DA/DC=BA/BC=35

→DA/DA+DC=3/3+5

→AD/AC=38

→AD=3/8AC=94

→BD=√AB2+AD2=9√54→BD=AB2+AD2=954

→sinˆABD=ADBD=1√5→sin⁡ABD^=ADBD=15

      cosˆABD=ABBD=2√5cos⁡ABD^=ABBD=25

      tanˆABD=sinˆABDcosˆABD=12tan⁡ABD^=sin⁡ABD^cos⁡ABD^=12

      cotˆABD=1tanˆABD=2

\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\)

\(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\)

Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk )

=> \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\)

\(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\)

\(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\)

\(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\)