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\(\sqrt{4x-2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}\)(với \(x\ge\dfrac{1}{4}\))
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left(\sqrt{4x-1}-1\right)+\left(\sqrt{4x-1}+1\right)\)
\(=2\sqrt{4x-1}\) (với \(x\ge\dfrac{1}{4}\))
Đặt \(D=\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\) (D >/ 0 với mọi 1/2 < x)
\(\Rightarrow D^2=2\sqrt{4x-1}-2\sqrt{4x^2-4x+1}=2\sqrt{4x-1}-2\left|2x-1\right|=2\sqrt{4x-1}-2\left(1-2x\right)=4x-2+2\sqrt{4x-1}\)
\(\Rightarrow D=\sqrt{D^2}=\sqrt{4x-2+2\sqrt{4x-1}}=\left|\sqrt{4x-1}+1\right|=\sqrt{4x-1}+1\)
Tớ làm nốt nè :3
\(1b.3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}=8\sqrt{2}\)
\(c.\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=4\)
\(2a.\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow4x^2-4x+1=9\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(b.\sqrt{4x-4}-\sqrt{9x-9}+5\sqrt{x-1}=7\left(x\ge1\right)\)
\(\Leftrightarrow2\sqrt{x-1}-3\sqrt{x-1}+5\sqrt{x-1}=7\)
\(\Leftrightarrow4\sqrt{x-1}=7\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{4}\)
\(\Leftrightarrow x=\dfrac{65}{16}\)
c. Sai đề.
\(A=\sqrt{2x-\sqrt{4x-1}}-\sqrt{2x+\sqrt{4x-1}}\)
\(A^2=\left(\sqrt{2x-\sqrt{4x-1}}-\sqrt{2x+\sqrt{4x-1}}\right)^2\)
\(A^2=2x-\sqrt{4x-1}+2x+\sqrt{4x-1}-2\sqrt{\left(2x-\sqrt{4x-1}\right)\left(2x+\sqrt{4x-1}\right)}\)
\(A^2=4x-2\sqrt{4x^2-4x+1}\)
\(A^2=4x-2\sqrt{\left(2x-1\right)^2}\)
\(A^2=4x-2\left|2x-1\right|\)
\(A^2=4x-2\left(1-2x\right)\) (vì\(\dfrac{1}{4}\le x\le\dfrac{1}{2}\)
\(A^2=8x-2\)
\(A=\sqrt{8x-2}\)
Cho tam giác ABC vuông tại A, AB<AC. Trung tuyến AM, ˆACB=x,ˆAMB=yACB^=x,AMB^=yChứng minh (sin x +cos x)2=1+sin y
MÌNH CẦN GẤP MỌI NGƯỜI GIÚP MÌNH NHA
\(=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}\)
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1\)
\(=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.\)
a) \(\sqrt{4x+8}-\sqrt{9x+18}+\sqrt{x+2}=\sqrt{x+5}\)
\(\Leftrightarrow\sqrt{4\left(x+2\right)}-\sqrt{9\left(x+2\right)}+\sqrt{x+2}=\sqrt{x+5}\)
\(\Leftrightarrow2\sqrt{x+2}-3\sqrt{x+2}+\sqrt{x+2}=\sqrt{x+5}\)
\(\Leftrightarrow0\sqrt{x+2}=\sqrt{x+5}\Leftrightarrow0=\sqrt{x+5}\)
\(\Leftrightarrow0=x+5\Leftrightarrow-5=x\)
Vậy phương trình đã cho có nghiệm duy nhất là x = -5
b) ĐKXĐ: \(x\ge0;x\ne1\)
\(T=\left(\dfrac{1}{1+2\sqrt{x}}-\dfrac{1}{\sqrt{3}+2}\right):\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)
\(=\left(\dfrac{\sqrt{3}+2-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(\sqrt{3}+2\right)}\right):\left(\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\right)\)
\(=\dfrac{1-2\sqrt{x}+\sqrt{3}}{\left(1+2\sqrt{x}\right)\left(\sqrt{3}+2\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
a) Bổ sung: ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x+2}XĐ\Leftrightarrow x+2\ge0\\\sqrt{x+5}XĐ\Leftrightarrow x+5\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ge-5\end{matrix}\right.\Rightarrow}x\ge-2}\) Sau khi tìm được x = -5 ta thấy k thỏa mãn Đk: \(x\ge-2\)
Vậy pt đã cho là vô nghiệm
\(D=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4x}{x-1}\)ĐK:x\(\ge\)0.x khác 0
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x}{x-1}\)
\(=\dfrac{4x+4\sqrt{x}}{x-1}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
Quy đồng lên ta được:
A=\(\sqrt{1+4X}\) - \(\sqrt{1-4X}\)