Giai phuong trinh
\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-2\sqrt{2x-5}}=2\sqrt{2}\)
Những câu hỏi liên quan
\(\sqrt{x+2+3\sqrt{ }2x-5}+\sqrt{x-2-\sqrt{ }2x-5}=2\sqrt{2}\) 2\(\sqrt{2}\) giai phuong trinh
giai phuong trinh: \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x-1}\)
giai phuong trinh: \(\sqrt[3]{x^2+4x+3}+\sqrt[3]{4x^2-9x-3}=\sqrt[3]{3x^2-2x+2}+\sqrt[3]{2x^2-3x-2}\)
giai phuong trinh \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\Leftrightarrow\sqrt{2x-5-6\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}+1\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}-3+\sqrt{2x-5}+1=4\\\sqrt{2x-5}-3+\sqrt{2x-5}+1=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}-2=4\\2\sqrt{2x-5}-2=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}=6\\2\sqrt{2x-5}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}=3\\\sqrt{2x-5}=-1\left(L\right)\end{cases}}\)
\(\Leftrightarrow2x-5=9\)
\(\Leftrightarrow x=7\)
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Giai phuong trinh :\(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x+8}=1+\sqrt{3}\)
giai phuong trinh : \(2x^2\left(5-\sqrt[3]{5x-x^3}\right)=2x^3+17x-8\)
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
Giai phuong trinh:
\(\sqrt{x^2+2x-2}+2\sqrt{x^2+2x-2}=x+2\)
Bình phương hai vế lên ta được:
x2+2x-2+4(x2+2x-2)=(x+2)2
<=>x2+2x-2+4x2+8x-8=x2+4x+4
<=>x2+4x2-x2+2x+8x-4x-2-8-4=0
<=>4x2+6x-14=0
<=>2x2+3x-7=0
Đến đây bạn tự làm tiếp nha. Nhớ k cho mk đấy
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Giai phuong trinh:
\(\sqrt{x^2+2x-2}+2\sqrt{x^2+2x-2}=x+2\)