Question

[Phan Trọng Hoàn]

làm nhanh hộ mk cái, mk đang cần gấp

Answer 1

[mystic and ma kết]

mình muốn âm điểm

Answer 2

[Hoàng Như Quỳnh]

\(a,x-\sqrt{x-4\sqrt{x}+4}=8\)

\(x-\sqrt{\left(\sqrt{x}-2\right)^2}=8\)

\(x-\left|\sqrt{x}-2\right|=8\)

\(TH1:0\le x\le2\)

\(x-2+\sqrt{x}=8\)

\(x+\sqrt{x}-10=0\)

\(\sqrt{\Delta}=1-\left(4.-10\right)=\sqrt{41}\)

\(\orbr{\begin{cases}x_1=\frac{\sqrt{41}-1}{2}\left(KTM\right)\\x_2=\frac{-\sqrt{41}-1}{2}\left(KTM\right)\end{cases}}\)

\(TH2:x>2\)

\(x-\sqrt{x}+2=8\)

\(x-\sqrt{x}-6=0\)

\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)

\(\orbr{\begin{cases}\sqrt{x}+2=0\\\sqrt{x}-3=0\end{cases}\orbr{\begin{cases}\sqrt{x}=-2\left(KTM\right)\\x=9\left(TM\right)\end{cases}}}\)

\(b,\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)

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

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

\(\left|\frac{1}{2}x+1\right|-\sqrt{5}+1=0\)

\(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)

\(\orbr{\begin{cases}\frac{1}{2}x+1=\sqrt{5}-1\\\frac{1}{2}x+1=1-\sqrt{5}\end{cases}\orbr{\begin{cases}\frac{1}{2}x=\sqrt{5}-2\\\frac{1}{2}x=-\sqrt{5}\end{cases}\orbr{\begin{cases}x=2\sqrt{5}-4\\x=-2\sqrt{5}\end{cases}}}}\)

\(c,\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

\(\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}+1\right|=4\)

\(\sqrt{2x-5}+3+\sqrt{2x-5}+1=4\)

\(\sqrt{2x-5}=0\)

\(x=\frac{5}{2}\left(TM\right)\)