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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: \(x\ge-1\)
\(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)
\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
a) Ta có: \(\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\left|\sqrt{3}+1\right|}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}\)(Vì \(\sqrt{3}>1>0\))
\(=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(a=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\left(3+\sqrt{3}\right)^2}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}\)
Xét \(A=\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}>0\)
\(A^2=6+2\sqrt{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}=6+2\sqrt{2}\)
\(\Rightarrow A=\sqrt{6+2\sqrt{2}}\)
\(\Rightarrow\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}=\sqrt{6+2\sqrt{2}}-\sqrt{6+2\sqrt{2}}=0\)
\(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\left(\sqrt{10}-\sqrt{2}\right)\) = \(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\left(\sqrt{10}-\sqrt{2}\right)\)
= \(2\sqrt{4+\sqrt{5}-1}\left(\sqrt{10}-\sqrt{2}\right)\) = \(\sqrt{2}\sqrt{6+2\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\)
= \(\sqrt{2}\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\) = \(\sqrt{2}\left(\sqrt{5}+1\right)\left(\sqrt{10}-\sqrt{2}\right)\)
= \(\sqrt{10}+\sqrt{2}\left(\sqrt{10}-\sqrt{2}\right)\) = \(10-2=8\)
dòng 2 : ta đưa nò về dạng hằng đẳng thức ; rồi đưa ra khỏi căn
tiếp là ta đưa \(\sqrt{2}\) vào để biến đổi thành hằng đẳng thức tiếp để bỏ căn
vì 2 = \(\sqrt{2}.\sqrt{2}\) nên ta đưa \(\sqrt{2}\) vào thì còn lại \(\sqrt{2}\) bênh ngoài
hiểu chưa
Bài 1: Ta có:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{6-2}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{6+2-2\sqrt{6.2}}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{(\sqrt{6}-\sqrt{2})^2}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2}=\frac{6-2}{2}=2\)
Bài 2:
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}\)
\(=16+2(\sqrt{20}-2)=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow A=\sqrt{10}+\sqrt{2}\)
1. \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+\sqrt{84}\)= -6,423305878
2. \(\sqrt{150}+\sqrt{1,6}\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)= 24,79207036
NHA 
Vũ Hoàng Thiên An ! ! !
K VÀ KB NHA !
1)
dat \(a=\sqrt[3]{x+1};b=\sqrt[3]{7-x}\)
ta co b=2-a
a^3+b^3=x+1+7-x=8
a^3+b^3=a^3+b^3+3ab(a+b)
ab(a+b)=0
suy ra a=0 hoac b=0 hoac a=-b
<=> x=-1; x=7
a=-b
a^3=-b^3
x+1=x+7 (vo li nen vo nghiem)
cau B tuong tu
2)
tat ca cac bai tap deu chung 1 dang do la
\(\sqrt[3]{a+m}+\sqrt[3]{b-m}\)voi m la tham so
dang nay co 2 cach
C1 lap phuong VD: \(B^3=10+3\sqrt[3]{< 5+2\sqrt{13}>< 5-2\sqrt{13}>}\left(B\right)\)
B^3=10-9B
B=1 cach nay nhanh nhung kho nhin
C2 dat an
\(a=\sqrt[3]{5+2\sqrt{13}};b=\sqrt[3]{5-2\sqrt{13}}\)
de thay B=a+b
a^3+b^3=10
ab=-3
B^3=10-9B
suy ra B=1
tuong tu giai cac cau con lai.
Bài 1:
a. Đặt \(a=\sqrt[3]{x+1}\); \(b=\sqrt[3]{7-x}\). Ta có:
\(\hept{\begin{cases}a+b=2\\a^3+b^3=8\end{cases}\Leftrightarrow a^3+\left(2-a\right)^3=8\Leftrightarrow...\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}a=0\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=0\\\sqrt[3]{7-x}=2\end{cases}}\)hoặc \(\hept{\begin{cases}\sqrt[3]{x+1}=2\\\sqrt[3]{7-x}=0\end{cases}}\)
\(\Leftrightarrow x=-1\)hoặc \(x=7\)
\(a\text{) }\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
\(b\text{) }\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\\ =\sqrt{18+3+2\sqrt{54}}-\sqrt{18+3-2\sqrt{54}}\\ =\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\\ =\sqrt{18}+\sqrt{3}-\sqrt{18}+\sqrt{3}\\ =2\sqrt{3}\)
\(d\text{) }\sqrt{x+1+2\sqrt{x}}\left(x\ge0\right)\\ =\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
\(e\text{) }\sqrt{2x+3+2\sqrt{x^2+3x+2}}\left(x\le-2;x\ge-1\right)\\ =\sqrt{\left(x+2\right)+\left(x+1\right)+2\sqrt{\left(x+1\right)\left(x+2\right)}}=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
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)\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{x}.1+1^2}=\sqrt{x}+1\)
=-3,068712231
3,068712231
K mình nha