Giai phuong trinh : \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
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giai phuong trinh
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
ĐK: \(x\ge-2\)
\(pt\Leftrightarrow\frac{x+5-\left(x+2\right)}{\sqrt{x+5}+\sqrt{x+2}}.\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)=3\)
\(\Leftrightarrow3.\frac{1+\sqrt{x+2}.\sqrt{x+5}}{\sqrt{x+2}+\sqrt{x+5}}=3\)
\(\Leftrightarrow1+\sqrt{x+2}\sqrt{x+5}=\sqrt{x+2}+\sqrt{x+5}\)
\(\Leftrightarrow\left(\sqrt{x+2}-1\right)\left(\sqrt{x+5}-1\right)=0\)
\(\Leftrightarrow\sqrt{x+2}=1\text{ hoặc }\sqrt{x+5}=1\)
\(\Leftrightarrow x=-1\text{ (nhận) hoặc }x=-4\text{ (loại)}\)
Vậy tập nghiệm của pt là: \(S=\left\{1\right\}\)
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giai phuong trinh : \(2x^2\left(5-\sqrt[3]{5x-x^3}\right)=2x^3+17x-8\)
giai phuong trinh
\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Leftrightarrow\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Rightarrow\left(x-2\right)^2=x^2-4\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow x=2\)
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Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{x-2}=b;\) ta có:
\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)
\(\Leftrightarrow2a^2+ab-2ab-b^2=0\)
⇔ \(\left(2a+b\right)\left(a-b\right)=0\)
⇔ \(\left[{}\begin{matrix}2\sqrt[3]{x+2}=-\sqrt[3]{x-2}\\\sqrt[3]{x-2}=\sqrt[3]{x+2}\end{matrix}\right.\)⇔ \(x=-\frac{14}{9}\)
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Giai phuong trinh va he phuong trinh:
a) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
b) \(x^2+3x+1=\left(x+3\right).\sqrt{x^2+1}\)
c) \(\left\{{}\begin{matrix}x^2+y^2=11\\x+xy+y=3+4\sqrt{2}\end{matrix}\right.\)
giai phuong trinh \(\sqrt{x\left(x-3\right)}-\sqrt{7x-3}=2\sqrt{x^2}\)
Giai phuong trinh
\(\sqrt{x.\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2x\)
x=0
nhanh nhất có thể
X=0
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giai he phuong trinh \(\left\{{}\begin{matrix}\sqrt{x+1}+\sqrt{y-1}=2+\sqrt{6}\\x+y=5+2\sqrt{6}\end{matrix}\right.\)
Giai phuong trinh:
\(28+\sqrt[3]{x^2}=3x+2\sqrt[3]{x}+\left(x-4\right)\sqrt{x-7}\)
Giai phuong trinh:
\(\sqrt{x+\frac{3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\)
\(DKXD:x>0\)
\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(\frac{1}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{1}{2\left(x+1\right)}\right)=0\)
\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\left(\text{ }x-1\right)\left(x^2+x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy PT có 2 nghiệm \(x=1;x=3\)
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