Rút gọn bt
P=\(\sqrt{x+4+2\sqrt{x+5}}-\sqrt{x+4-2\sqrt{x+5}}vớix\ge-4\)
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\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Học tốt
Bài làm:
a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(B=2+\sqrt{5}-\sqrt{5}+2\)
\(B=4\)
Bài làm:
a) \(\sqrt{x^2+6x+9}=4\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)
\(\Leftrightarrow\left|x+3\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=4\\x+3=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-7\end{cases}}\)
b) \(\sqrt{x^2+4x+4}=x-3\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=x-3\)
\(\Leftrightarrow\left|x+2\right|=x-3\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=x-3\\x+2=3-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x=-5\left(vl\right)\\2x=1\end{cases}}\Rightarrow x=\frac{1}{2}\)
cách khác
a,\(\sqrt{x^2+6x+9}=4\)
\(< =>x^2+6x+9=16\)
\(< =>x^2-x+7x-7=0\)
\(< =>x\left(x-1\right)+7\left(x-1\right)=0\)
\(< =>\left(x+7\right)\left(x-1\right)=0\)
\(< =>\orbr{\begin{cases}x=-7\\x=1\end{cases}}\)
b,\(\sqrt{x^2+4x+4}=x-3\)
\(< =>x^2+4x+4=x^2-6x+9\)
\(< =>x^2+4x+4-x^2+6x-9=0\)
\(< =>10x-5=0< =>x=\frac{1}{2}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
a. \(11+2\sqrt{10}=\left(\sqrt{10}+1\right)^2\)
b. \(12-2\sqrt{11}=\left(\sqrt{11}-1\right)^2\)
c.\(23+2\sqrt{22}=\left(\sqrt{22}+1\right)^2\)
Bài làm:
a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b) \(4x^2-5=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)
c) \(3x^2-1=\left(x\sqrt{3}-1\right)\left(x\sqrt{3}+1\right)\)
d) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
e) \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)
a) ĐKXĐ: thỏa mãn với mọi a thực
b) ĐKXĐ: \(\frac{1}{2a+1}>0\)
\(\Rightarrow2a+1>0\Rightarrow2a>-1\Leftrightarrow a>-\frac{1}{2}\)
c) ĐKXĐ: \(a\left(1-a\right)\ge0\)
+ Nếu: \(\hept{\begin{cases}a\ge0\\1-a\ge0\end{cases}}\Leftrightarrow1\ge a\ge0\)
+ Nếu: \(\hept{\begin{cases}a\le0\\1-a\le0\end{cases}\Rightarrow}\hept{\begin{cases}a\le0\\a\ge1\end{cases}}\)(vô lý)
Vậy \(0\le a\le1\)
d) ĐKXĐ: \(\frac{2}{\left(a-2\right)\left(a+3\right)}>0\)
\(\Rightarrow\left(a-2\right)\left(a+3\right)>0\)
+ Nếu: \(\hept{\begin{cases}a-2>0\\a+3>0\end{cases}}\Rightarrow a>2\)
+ Nếu: \(\hept{\begin{cases}a-2< 0\\a+3< 0\end{cases}}\Rightarrow a< -3\)
Vậy \(\orbr{\begin{cases}a>2\\a< -3\end{cases}}\)
Để biểu thức có nghĩa thì :
\(\sqrt{4+a^2}\left(đk:\forall a-tmđk\right)\)
\(\sqrt{\frac{1}{2a+1}}\left(đk:a\ne-\frac{1}{2};a\ge-\frac{1}{2}\Leftrightarrow a>-\frac{1}{2}\right)\)
\(\sqrt{a\left(1-a\right)}\left(đk:a\ge0\right)\)
\(\sqrt{\frac{2}{\left(a-2\right)\left(a+3\right)}}\left(đk:a\ge2;a\ne2\Leftrightarrow a>2\right)\)
Sửa đề :
\(P=\sqrt{x+5+2\sqrt{x+4}}-\sqrt{x+5-2\sqrt{x+4}}\)\(\left(x\ge-4\right)\)
\(=\sqrt{\left(x+4\right)+2\sqrt{x+4}+1}-\sqrt{\left(x+4\right)-2\sqrt{x+4}+1}\)
\(=\sqrt{\left(\sqrt{x+4}+1\right)^2}-\sqrt{\left(\sqrt{x+4}-1\right)^2}\)
\(=\left|\sqrt{x+4}+1\right|-\left|\sqrt{x+4}-1\right|\)
\(=\sqrt{x+4}+1-\sqrt{x+4}+1=2\)
Vậy \(P=2\)
Tại sao lại phải sửa đề ạ?