Câu a:
TH1: \(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x+2x-1=2\Leftrightarrow x=1\)
TH2:\(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x-2x+1=2\Leftrightarrow x=-1\)
ĐK: \(x\le2\)
\(x+\sqrt{4x^2-4x+1}=2\)
\(\Leftrightarrow\)\(\sqrt{4x^2-4x+1}=2-x\)
\(\Leftrightarrow\)\(4x^2-4x+1=4-4x+x^2\)
\(\Leftrightarrow\)\(3x^2=3\)
\(\Leftrightarrow\)\(x=\pm1\)(t/m)
Vậy...
\(1-\sqrt{4x^2-20x+25}=0\)
\(\Leftrightarrow\)\(\sqrt{4x^2-20x+25}=1\)
\(\Leftrightarrow\)\(4x^2-20x+24=0\)
\(\Leftrightarrow\)\(x^2-5x+6=0\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy...
a) \(\sqrt{4x^2-4x+1}=2-x\left(x\le2\right)\)
<=> \(4x^2-4x+1=\left(2-x\right)^2\)
<=> \(3x^2-3=0\)
<=> \(x_1=1_{ }\)(TM) ; \(x_2=-1_{ }_{ }\)(TM)
b)\(1-\sqrt{4x^4-20x^2+25}=0\)
Đặt x2 = a ( a\(\ge\)0)
=> pt có dạng :
\(1-\sqrt{4a^2-20a+25}=0\)
<=> \(\sqrt{4a^2-20a+25}=1\)
<=> \(4a^2-20a+25=1\)
=> \(a_1=3\)\(\left(TM\right)\) ; \(a_2=2\left(TM\right)\)
\(a_1=3\Rightarrow x=\pm\sqrt{3}\)
\(a_2=2\Rightarrow x=\pm\sqrt{2}\)
ĐK : \(x\ge\frac{1}{2}\)
\(x+\sqrt{4x^2-4x+1}=2\)
\(x+\sqrt{\left(2x-1\right)^2}=2\)
\(x+2x-1=2\)
\(3x=3\)
\(x=1\left(TM\right)\)
Đk : \(x\ge\sqrt{\frac{5}{2}}\)
\(1-\sqrt{4x^4-20x^2+25}=0\)
\(1-\sqrt{\left(2x^2-5\right)^2}=0\)
\(1-2x^2+5=0\)
\(2x^2=6\)
\(x^2=3\)
\(x=\pm\sqrt{3}\)
xin lỗi câu 2 mk đọc nhầm đề
\(1-\sqrt{4x^4-20x^2+25}=0\)
\(\Leftrightarrow\)\(\sqrt{4x^4-20x^2+25}=1\)
\(\Leftrightarrow\)\(4x^4-20x^2+24=0\)
\(\Leftrightarrow\)\(x^4-5x^2+6=0\)
\(\Leftrightarrow\)\(\left(x^2-2\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\)\(\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+\sqrt{3}\right)=0\)
đến đây tự làm
