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\(a\text{) }\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
\(b\text{) }\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\\ =\sqrt{18+3+2\sqrt{54}}-\sqrt{18+3-2\sqrt{54}}\\ =\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\\ =\sqrt{18}+\sqrt{3}-\sqrt{18}+\sqrt{3}\\ =2\sqrt{3}\)
\(d\text{) }\sqrt{x+1+2\sqrt{x}}\left(x\ge0\right)\\ =\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
\(e\text{) }\sqrt{2x+3+2\sqrt{x^2+3x+2}}\left(x\le-2;x\ge-1\right)\\ =\sqrt{\left(x+2\right)+\left(x+1\right)+2\sqrt{\left(x+1\right)\left(x+2\right)}}=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
Xem lại đề câu c nha.
a)\(\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
b)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
=\(\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\sqrt{3^2}}\)
=\(\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
=\(3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}\)
=\(2\sqrt{3}\)
c)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
ÁP dụng HĐT \(\sqrt{a+b}\pm\sqrt{a-b}=\sqrt{2\left(a.\sqrt{a^2\pm b}\right)}\)ta có:
=\(\sqrt{2\left(4+\sqrt{4^2-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{16-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{6-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}-1\right)^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{5}-1\right)}\)
=\(\sqrt{2\left(3+\sqrt{5}\right)}\)
=\(\sqrt{6+\sqrt{5}}=\sqrt{5}+1\)
d)\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{x}.1+1^2}=\sqrt{x}+1\)
chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
1/Em không chắc nha, nhất là câu c ý, nó sai sai hay là em làm sai nhỉ?
a) ĐK \(x\ge0\). Bình phương hai vế:
\(x+5=x+2\sqrt{x}+1\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\) (TMĐK)
b)ĐK \(0\le x\le1\) . Bình phương hai vế:
\(2\sqrt{x\left(1-x\right)}=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(TMĐK\right)\)
c) ĐK: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\Leftrightarrow5\le x\le3\) (vô lí))
Vậy không tồn tại x thỏa mãn đề bài.
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
a) \(A=\sqrt{10+\sqrt{99}}=\sqrt{10+3\sqrt{11}}=\frac{1}{\sqrt{2}}.\sqrt{20+6\sqrt{11}}\)
\(=\frac{1}{\sqrt{2}}.\sqrt{\left(3+\sqrt{11}\right)^2}=\frac{3+\sqrt{11}}{2}\)
b) \(B=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
c) bn ktra lại đề
d) ĐK: \(x\ge0\)
\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
e) đk: \(x\ge-1\)
\(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)
\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)