\(x^2\cdot\sqrt{7}=\dfrac{\sqrt{6363}}{\sqrt{707}}\)
Những câu hỏi liên quan
Bài 1
a) x2 . \(\sqrt{7}\) =\(\frac{\sqrt{6363}}{\sqrt{707}}\)
b) x =\(\sqrt{2x-1}\)
Bài 2
Cho M =(2-\(\frac{x-\sqrt{x}}{\sqrt{x}-1}\)).(2+\(\frac{x+\sqrt{x}}{\sqrt{x}+1}\)) với x ≥ 0; x ≠ 1
a) Rút gọn M
b)Đặt P = M +\(\sqrt{x}\). Tìm GTLN của P.
https://i.imgur.com/59DavAU.jpg
Mấy cái này chỉ đơn giản là sử dụng các phép biến đổi đơn giản của biểu thức chứa căn bậc hai thôi nên bạn chú ý xem lại các bài trong SGK là làm được rồi! Chúc bạn học tốt nhé! 
a) \(x^2.\sqrt{7}=\frac{\sqrt{6363}}{\sqrt{707}}=\sqrt{9}=3\)
\(\Rightarrow x^2=\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}\Rightarrow x=\)\(\left[{}\begin{matrix}-\sqrt{\frac{3\sqrt{7}}{7}}\\\sqrt{\frac{3\sqrt{7}}{7}}\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge\frac{1}{2}\)
PT \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=2x-1\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\end{matrix}\right.\)(thỏa mãn)
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khử mẫu bt lấy căn :
a) \(3xy\cdot\sqrt{\dfrac{2}{xy}}\)
b)\(x\cdot\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}\)
c) \(xy\cdot\sqrt{\dfrac{1}{xy}}+x\cdot\sqrt{\dfrac{y}{x}}-y\cdot\sqrt{\dfrac{x}{y}}\)
a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)
b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)
\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)
c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)
\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)
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Chứng minh biểu thức không phụ thuộc vào x
\(K=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}\cdot\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}\cdot\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(K=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)
Vậy K không phụ thuộc vào x
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rút gọnAleft(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}Bsqrt{6+2sqrt{5}}-sqrt{6-2sqrt{5}}Cleft(sqrt{7}-sqrt{10}right)^2+sqrt{280}Ddfrac{sqrt{99}}{sqrt{11}}+sqrt{7}cdotsqrt{63}-sqrt{sqrt{81}}Esqrt{27}left(s-sqrt{5}right)^2cdotleft(3sqrt{48}right)giải chi tiết ra giúp mik nha,cảm ơn nhiều
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rút gọn
A=\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
B=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
C=\(\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}\)
D=\(\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}\)
E=\(\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)\)
giải chi tiết ra giúp mik nha,cảm ơn nhiều
Rút gọn:
\(A=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]\cdot\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{xy^3}+\sqrt{x^3y}}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
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bài 1 tính
a, sqrt{left(1-sqrt{5}right)^2}+1
b, sqrt{3+2cdotsqrt{2}}-2
c, sqrt{b^2-b+dfrac{1}{4}}-left(2b-dfrac{1}{2}right)left(vsbgedfrac{1}{2}right)
d, sqrt{7+2cdotsqrt{10}}
e. sqrt{11-4cdotsqrt{7}}
f, sqrt{x-2cdotsqrt{x-1}}
g, 3x+sqrt{x^2-2x+1}
h, sqrt{y+2sqrt{y^2-2y+1}} (voi y1)
i, sqrt{17-2sqrt{32}}+sqrt{17+2sqrt{32}}
k, sqrt{5+2sqrt{6}}-sqrt{5-2sqrt{6}}
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bài 1 tính
a, \(\sqrt{\left(1-\sqrt{5}\right)^2}+1\)
b, \(\sqrt{3+2\cdot\sqrt{2}}-2\)
c, \(\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)\left(vsb\ge\dfrac{1}{2}\right)\)
d, \(\sqrt{7+2\cdot\sqrt{10}}\)
e. \(\sqrt{11-4\cdot\sqrt{7}}\)
f, \(\sqrt{x-2\cdot\sqrt{x-1}}\)
g, \(3x+\sqrt{x^2-2x+1}\)
h, \(\sqrt{y+2\sqrt{y^2-2y+1}}\) (voi y>1)
i, \(\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
k, \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(a.\sqrt{\left(1-\sqrt{5}\right)^2}+1=\left|1-\sqrt{5}\right|+1=\sqrt{5}-1+1=\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}-2=\sqrt{\left(\sqrt{2}+1\right)^2}-2=\sqrt{2}+1-2=\sqrt{2}-1\)
\(c.\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)=\sqrt{\left(b-\dfrac{1}{2}\right)^2}-2b+\dfrac{1}{2}=b-\dfrac{1}{2}-2b+\dfrac{1}{2}=-2b\)
\(d.\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}\)
\(e.\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-2\)
\(g.3x+\sqrt{x^2-2x+1}=3x+\sqrt{\left(x-1\right)^2}\)
* \(x\ge1\Rightarrow3x+\left|x-1\right|=3x+x-1=4x-1\)
* \(x< 1\Rightarrow3x+\left|x-1\right|=3x+1-x=2x+1\)
\(h.\sqrt{y+2\sqrt{y^2-2y+1}}=\sqrt{y+2\sqrt{\left(y-1\right)^2}}=\sqrt{y+2y-2}=\sqrt{3y-2}\left(y\ge1\right)\) hoặc: \(\sqrt{y+2-2y}=\sqrt{-y+2}\left(y< 1\right)\)
\(H=\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
\(H^2=17-2\sqrt{32}+17+2\sqrt{32}+2\sqrt{\left(17-2\sqrt{32}\right)\left(17+2\sqrt{32}\right)}=34+2\sqrt{161}\)
\(H=\sqrt{34+2\sqrt{161}}\)
\(k.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
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Giải các phương trình sau
a) -x^2+4cdot x+12cdotsqrt{2cdot x+1}
b) x+sqrt{x+dfrac{1}{2}+sqrt{x+dfrac{1}{4}}}2
c) 5cdot x^2-2cdot x+1left(4cdot x-1right)cdotsqrt{x^2+1}
d) left(2cdot x-1right)cdotsqrt{10-4cdot x^2}5-2cdot x
e) sqrt{2cdot x-1}-sqrt{x+1}2cdot x-4
f) sqrt{x^2-2cdot x}+sqrt{2cdot x^2+4cdot x}2cdot x
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Giải các phương trình sau
a) \(-x^2+4\cdot x+1=2\cdot\sqrt{2\cdot x+1}\)
b) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
c) \(5\cdot x^2-2\cdot x+1=\left(4\cdot x-1\right)\cdot\sqrt{x^2+1}\)
d) \(\left(2\cdot x-1\right)\cdot\sqrt{10-4\cdot x^2}=5-2\cdot x\)
e) \(\sqrt{2\cdot x-1}-\sqrt{x+1}=2\cdot x-4\)
f) \(\sqrt{x^2-2\cdot x}+\sqrt{2\cdot x^2+4\cdot x}=2\cdot x\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
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\(\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}\cdot\left(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt[]{xy}}\right)\)
Thu gọn biểu thức trên
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
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Tìm x, biết:
\(x=\dfrac{19}{11}\cdot\dfrac{5}{14}+\dfrac{1}{11}\cdot\dfrac{5}{7}-\sqrt{\dfrac{25}{4}}\cdot\dfrac{2}{11}\)