\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
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H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
NT
3
22 tháng 6 2019
\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)
Tương tự \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\); \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)
TP
23 tháng 6 2019
\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}-4\)
HP
31 tháng 8 2021
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
24 tháng 8 2021
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
31 tháng 10 2021
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
24 tháng 8 2021
`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`
`=2/sqrt2=sqrt2`
`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`
`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`
`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`
`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`
`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`
`=(-2sqrt3)/sqrt2=-sqrt6`
`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`
`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`
`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`
`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`
`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`
`=(2sqrt3)/sqrt2=sqrt6`
`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`
`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`
`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`
`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`
`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`
`=2/sqrt2=sqrt2`
25 tháng 8 2021
a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)
b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2021
ĐKXĐ: \(a^3+a^2-3\ge0\) (1)
Đặt \(\sqrt{a^3+a^2-3}=x\ge0\Rightarrow a^3+a^2+4=x^2+7\)
Phương trình trở thành:
\(\sqrt{x^2+7}+x=7\)
\(\Leftrightarrow\sqrt{x^2+7}=7-x\) (\(x\le7\))
\(\Leftrightarrow x^2+7=x^2-14x+49\)
\(\Rightarrow14x=42\)
\(\Rightarrow x=3\)
\(\Rightarrow a^3+a^2-3=9\)
\(\Rightarrow a^3+a^2-12=0\)
\(\Rightarrow\left(a-2\right)\left(a^2+3a+6\right)=0\)
\(\Rightarrow a=2\) (thỏa mãn (1))
29 tháng 7 2020
\(A=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
\(A=\sqrt{9+6\sqrt{5}+5}+\sqrt{9-6\sqrt{5}+5}\)
\(A=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(A=3+\sqrt{5}+3-\sqrt{5}=6\)
b) \(B=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(B=\sqrt{3-4\sqrt{3}+4}-\sqrt{3+4\sqrt{3}+4}\)
\(B=\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(B=2-\sqrt{3}-\sqrt{3}-2=-2\sqrt{3}\)
29 tháng 7 2020
Câu a tách 14 thành 5+9 . Có hằng đẳng thức
Câu b tương tự tách 7 thành 4+ 3 nhé
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+2+2-\sqrt{3}\)
\(=4\)
Còn cách nữa là bình phương
Đag làm thì ấn nhầm trả lời .V
Cách bình phương đây
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(\Rightarrow B^2=7+4\sqrt{3}+2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+7-4\sqrt{3}\)
\(=14+2\sqrt{49-48}\)
\(=14+2\)
\(=16\)
\(\Rightarrow B=\sqrt{16}=4\)