a/ \(\frac{x-2}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}+2}\)

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

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

b/ \(\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}=\frac{4+2\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}-4}{4+2\sqrt{3}+2\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\frac{4+2\sqrt{3}+\sqrt{3}+1-4}{4+2\sqrt{3}+2\sqrt{3}+2}=\frac{1+3\sqrt{3}}{6+4\sqrt{3}}\)

câu c nữa bạn

\(\sqrt{\left(1+x^2\right)^3}-4x^3=1-3x^4\)

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

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

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

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

Ta có:  \(\frac{\left(1+x^2\right)^2+\left(1+x^2\right)+1}{\sqrt{\left(1+x^2\right)^3}+1}-4x+3x^2\ge3x^2-4x+\frac{3}{2}>0\)

\(\Rightarrow x=0\)

Đề sai. Nếu a=2;b=1;c=0 thì \(a^4+b^4+c^4=16+1+0=17\)

\(a^2+b^2+c^2=4+1+0=5\)

4 đó bạn

HỌC TỐT!

Vậy x bằng ...

\(4^{x+x}\cdot4^{\sqrt{x+2}}+2^{x^3}=4^x\cdot4^{\sqrt{x+2}}+2^{x^3}\cdot2^{4\cdot x}:2^4\)

\(4^x\cdot4^x\cdot4^{\sqrt{x+2}}+2^{x^3}=4^x\cdot4^{\sqrt{x+2}}+2^{x^3}\cdot2^{4\cdot x}:16\)

\(4^x=2^{4\cdot x}:16\)

\(16=\frac{\left(2^4\right)^x}{4^x}\)

\(16=\frac{\left(2^4\right)^x}{4^x}\)

\(16=\frac{16^x}{4^x}\)

\(16=\left(\frac{16}{4}\right)^x\)

\(16=4^x\)

\(4^x=16\)

\(4^x=4^2\)

\(\Rightarrow x=2\)

Từ hàng thứ 2 qua thứ 3 là do cách triệt số khi chuyển vế 

Mình bổ sung nha:

\(4^x\cdot4^x\cdot4^{\sqrt{x+2}}+2^{x^3}=4^x\cdot4^{\sqrt{x+2}}+2^{x^3}\cdot2^{4\cdot x}:16\)

\(\frac{4^x\cdot4^{\sqrt{x+2}}}{4^x\cdot4^{\sqrt{x+2}}}+2^{x^3}-2^{x^3}=\cdot2^{4\cdot x}:16:4^x\)