Tìm x biết: \(\sqrt{x+3+14\cdot\sqrt{x-1}}\sqrt{x+8-6\cdot\sqrt{x-1}}=5\)
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Giải phương trình:a)sqrt[3]{14-x^3}+x2cdotleft(1+sqrt{x^2-2x-1}right)b) 5-3xleft(-125x^2+150x-41right)cdotsqrt{1-x^2}c)sqrt{2x^2+1}+sqrt{x^2+3x+2}sqrt{x^2-x+4}+sqrt{2x^2+2x+3}d) sqrt{x^2+15}+2sqrt{x^2+8}+3xe) sqrt{2x^4+2}cdotleft(sqrt{2-x}-sqrt{x}right)left(1-xright)cdotleft(x^2+1right)f) sqrt[3]{2037-x}-sqrt{x-2009}x^2-2009x-2008
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Giải phương trình:
a)\(\sqrt[3]{14-x^3}+x=2\cdot\left(1+\sqrt{x^2-2x-1}\right)\)
b) \(5-3x=\left(-125x^2+150x-41\right)\cdot\sqrt{1-x^2}\)
c)\(\sqrt{2x^2+1}+\sqrt{x^2+3x+2}=\sqrt{x^2-x+4}+\sqrt{2x^2+2x+3}\)
d) \(\sqrt{x^2+15}+2=\sqrt{x^2+8}+3x\)
e) \(\sqrt{2x^4+2}\cdot\left(\sqrt{2-x}-\sqrt{x}\right)=\left(1-x\right)\cdot\left(x^2+1\right)\)
f) \(\sqrt[3]{2037-x}-\sqrt{x-2009}=x^2-2009x-2008\)
giải bài nào hộ mk cx được ko cần lm hết đâu :) :) :)
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Tìm x, biết:
\(x=\dfrac{19}{11}\cdot\dfrac{5}{14}+\dfrac{1}{11}\cdot\dfrac{5}{7}-\sqrt{\dfrac{25}{4}}\cdot\dfrac{2}{11}\)
\(\sqrt{2\cdot x^2+4\cdot x+6}\) +\(\sqrt{3\cdot x^2+6\cdot x+12}\)=5-\(2\cdot x\)-\(x^2\)
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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Giải các phương trình sau:
a)\(\sqrt[3]{9-x}+\sqrt[3]{7+x}=4\)
b)\(\sqrt{x-1}\cdot\sqrt[4]{x^2-4}=\sqrt{x-2}\cdot\sqrt[4]{x^2-1}\)
c)\(\sqrt[4]{9-x^2}+\sqrt{x^2-1}-2\sqrt{2}=\sqrt[6]{x-3}\)
a) Áp dụng bđt AM-GM có:
\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)
\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)
Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)
Vậy...
b)Đk:\(x\ge2\)
Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)
Do \(x\ge2\Rightarrow x-1>0\)
Chia cả hai vế của pt cho x-1 ta được:
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy S={2}
c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)
Thay x=3 vào pt thấy thỏa mãn
Vậy S={3}
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GPT : x = \(\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
Giải pt : \(x=\sqrt{3-x}\cdot\sqrt{4-x}+\sqrt{4-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{3-x}\)
giải pt
\(x=\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
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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