a) ĐK: \(0\le x\le\frac{\sqrt{5}+1}{2}\)

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

\(\Leftrightarrow1-\sqrt{x^2-x}=\left(\sqrt{x}-1\right)^2\left(x\ge1\right)\)

\(\Leftrightarrow1-\sqrt{x^2-x}=x-2\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x\left(x-1\right)}=2\sqrt{x}-x\)

\(\Leftrightarrow\sqrt{x\left(x-1\right)}=\sqrt{x}\left(2-\sqrt{x}\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x-1}+\sqrt{x}-2=0\end{cases}}\)

TH1: x = 0 (Loại)

TH2: \(\sqrt{x-1}+\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x-1}=2-\sqrt{x}\)

\(\Leftrightarrow x-1=4-4\sqrt{x}+x\left(x\le4\right)\)

\(\Leftrightarrow4\sqrt{x}=5\Leftrightarrow\sqrt{x}=\frac{5}{4}\Leftrightarrow x=\frac{25}{16}\left(tm\right)\)

b) \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

ĐK: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{\left(x+1\right)\left(2x+6\right)}+\sqrt{\left(x+1\right)\left(x-1\right)}=2\left(x+1\right)\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{2x+6}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{2x+6}+\sqrt{x-1}-2\sqrt{x+1}=0\end{cases}}\)

TH1: \(\sqrt{x+1}=0\Leftrightarrow x=-1\left(l\right)\)

TH2: \(\sqrt{2x+6}=2\sqrt{x+1}-\sqrt{x-1}\)

\(\Leftrightarrow2x+6=4\left(x+1\right)+\left(x-1\right)-4\sqrt{x^2-1}\)

\(\Leftrightarrow2x+6=5x+3-4\sqrt{x^2-1}\)

\(\Leftrightarrow4\sqrt{x^2-1}=3x-3\Leftrightarrow16\left(x^2-1\right)=9x^2-18x+9\left(x\ge1\right)\)

\(\Leftrightarrow7x^2+18x-25=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-\frac{25}{7}\left(l\right)\end{cases}}\)

dk tu xd \(\sqrt{2x^2+8x+6}\) \(+\sqrt{x^2-1}=2x+2\)

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

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

đến đây bn tự giải nhé

1. a) Để hàm số đồng biến thì m-1>0\(\Rightarrow\)m>1  b) Để hàm số nghịch biến m-1<0\(\Rightarrow\)m<1        2. a) Tự làm  b) Xét phương trình hoành độ -2x+1=2x\(\Rightarrow\)x=1/4\(\Rightarrow\)   y=1/2. Vậy giao điểm của d và d' có tọa độ (1/4; 1/2)              

3 a)ĐKXĐ \(x\ge0\)\(x\ne1\)A=\(\frac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{-2}{\sqrt{x}+1}\)  b)Khi x= \(6-2\sqrt{5}\)thì A=\(\frac{-2}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\)=\(\frac{2}{\sqrt{5}}\)

Đề sai rùi 

\(R=\frac{\sqrt{\left(-\frac{2}{5}\right)^5.\left(-\frac{5}{8}\right)^3.5^2}}{\sqrt[3]{\left(-\frac{3}{4}\right)^3.\left(-\frac{5}{24}\right)^2.\left(-\frac{5}{3}\right)^4}}\)

\(=\frac{\sqrt{\frac{2^5}{5^5}.\frac{5^3}{8^3}.5^2}}{-\sqrt[3]{\frac{3^3}{4^3}.\frac{5^2}{24^2}.\frac{5^4}{3^4}}}\)

\(=\frac{\sqrt{\frac{1}{16}}}{-\sqrt[3]{\frac{1}{27}.5^6.\frac{1}{2^{12}}}}=\frac{\frac{1}{4}}{-\frac{1}{3}.5^2.\frac{1}{16}}=-\frac{12}{25}\)

1+90+5 48+5 345 43

ab+bc+ac=1 nha mn.. giúp với. 

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