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\(\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)=\(\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}=\sqrt{a-2}+2+2-\sqrt{a-2}=4\) (do2<=a<=4)
Bài 1:
a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)
\(=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\)
\(=\sqrt{2}\)
b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2-\sqrt{5}\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2\)
Bài 2:
a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne\pm2\)
Với \(x\ne\pm2\), ta có:
\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)
\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)
\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)
\(\Rightarrow x^2-4=8-x\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\) (TM)
Vậy phương trình có tập nghiệm là: S ={3; -4}
\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
1) pt có 2 dấu bằng.......t bỏ =1 được hong?
ĐK: \(\left\{{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}\le x\\x\ge1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}2x-1\le x^2\\x\ge1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x^2-2x+1\ge0\\x\ge1\end{matrix}\right.\Leftrightarrow}x\ge1}\)
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\Leftrightarrow x-2\sqrt{x-1}=x-1\Leftrightarrow4x-4=1\Leftrightarrow x=\dfrac{5}{4}\left(N\right)\)
Kl: x= 5/4
2) \(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=\sqrt{\left(a-2\right)+2\cdot2\cdot\sqrt{a-2}+4}+\sqrt{\left(a-2\right)-2\cdot2\cdot\sqrt{a-2}+4}=\sqrt{\left(a-2+2\right)^2}+\sqrt{\left(a-2-2\right)^2}=a+a-4=2a-4\)
chép lại cái đk, ghét nhất cái trò này của H24!! Viết đã đời cuối cùng công cốc !!
\(\left\{{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}\le x\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-4\le x^2\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-4x+4\ge0\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)