Giải phương trình \(\sqrt{x-2+\sqrt{2\cdot x+5}}+\sqrt{x+2+3\cdot\sqrt{2\cdot x-5}}=7\cdot\sqrt{2}\)
Những câu hỏi liên quan
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
Đọc tiếp
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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Bình luận (3)
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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Bình luận (1)
giải pt
\(x=\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
Giải phương trình:
a)\(\left(x+2\right)\cdot\left(x+4\right)+5\cdot\left(x+2\right)\cdot\sqrt{\frac{x+4}{x+2}}=6\)
b)\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
GPT : x = \(\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
\(\sqrt{2\cdot x^2+4\cdot x+6}\) +\(\sqrt{3\cdot x^2+6\cdot x+12}\)=5-\(2\cdot x\)-\(x^2\)
giải phương trình
\(\left(3-x\right)\cdot\sqrt{\left(3+x\right)\cdot\left(9+x^2\right)}=4\sqrt{5\cdot\left(3-x\right)}\)
giải phuong trình \(\left(3+x\right)\cdot\sqrt{\left(3+x\right)\cdot\left(9+x^2\right)}=4\cdot\sqrt{5\cdot\left(3-x\right)}\)