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a.
\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)
\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)
b.
\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)
\(\Leftrightarrow x^2-8=5x+1\)
\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)
............................
tương tự ..
c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)
=>x-5=0 hoặc x+5=1
=>x=-4 hoặc x=5
d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=7/2 hoặc x=-3/2
e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc 3 căn x+2=1
=>x=2 hoặc x+2=1/9
=>x=-17/9 hoặc x=2
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
đề sai r,,,,,,cái kia phải là x^2-x+1 chứ
nếu đúng như tôi thì bạn chỉ cần cho cái 2 vào trong căn rồi nhân liên hợp là ok
\(\sqrt{x^2+3}+\frac{4x}{\sqrt{x^2+3}}=5\sqrt{x}\)
\(\frac{\sqrt{x^2+3}.\sqrt{x^2+3}}{\sqrt{x^2+3}}+\frac{4x}{\sqrt{x^2+3}}=5\sqrt{x}\)
\(\frac{\sqrt{\left(x^2+3\right)\left(x^2+3\right)}}{\sqrt{x^2+3}}+\frac{4x}{\sqrt{x^2+3}}=\frac{5\sqrt{x.\left(x^2+3\right)}}{\sqrt{x^2+3}}\)
\(\frac{\sqrt{\left(x^2+3\right)^2}}{\sqrt{x^2+3}}+\frac{4x}{\sqrt{x^2+3}}=\frac{3\sqrt{x\left(x^2+3\right)}}{\sqrt{x^2+3}}\)
\(\Leftrightarrow\)\(x^2+4x+3=3\sqrt{x\left(x^2+3\right)}\)
\(\Leftrightarrow\left(x^2+4x+3\right)^2=\left[3\sqrt{x\left(x^2+3\right)}\right]^2\)
\(\Leftrightarrow x^4+8x^3+9=9.\left(x^3+3x\right)\)
\(\Leftrightarrow x^4+8x^3+9=9x^3+27x\)
\(\Leftrightarrow x^4+8x^3-9x^3-27x+9=0\)
\(\Leftrightarrow x^4-x^3-27x+9=0\)
\(\Leftrightarrow\left(x^4-27x\right)-\left(x^3-9\right)=0\)
\(\Leftrightarrow x\left(x^3-27\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x^2+3x+9\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left[x\left(x^2+3x+9\right)-x-3\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left[x^3+3x^2+9x-x-3\right]=0\)
\(\Leftrightarrow\left(x-3\right).\left[x^3+3x^2+8x-3\right]=0\)
Sao ở vế phải đang từ \(\frac{5\sqrt{x\left(x^2+3\right)}}{x^2+3}\) lại thành \(\frac{3\sqrt{x\left(x^2+3\right)}}{x^2+3}\)