giải pt sau: \(\sqrt{x^3+8}\) +x = \(\dfrac{2}{3}\) . (x2 +5)
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Y
24 tháng 6 2023
\(\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6+4\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{1;6+4\sqrt{5}\right\}\)
PN
1
25 tháng 7 2018
ĐKXĐ: \(x>4\)
\(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow\)\((\dfrac{\sqrt{x+5}}{\sqrt{x-4}})^2=(\dfrac{\sqrt{x-2}}{\sqrt{x+3}})^2\)
\(\Leftrightarrow\dfrac{x+5}{x-4}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\dfrac{x+5}{x-4}-\dfrac{x-2}{x+3}=0\)
\(\Leftrightarrow\dfrac{(x+5)\left(x+3\right)-\left(x-2\right)\left(x-4\right)}{(x-4)\left(x+3\right)}=0\)
\(\Leftrightarrow(x+5)\left(x+3\right)-\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow x^2+8x+15-x^2+6x-8=0\)
\(\Leftrightarrow14x-7=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2021
ĐKXĐ: \(x\ne-1\)
\(\dfrac{6x^2+4x+8}{x+1}=5\sqrt{2x^2+3}\)
\(\Rightarrow6x^2+4x+8=5\left(x+1\right)\sqrt{2x^2+3}\)
\(\Leftrightarrow2\left(2x^2+3\right)-5\left(x+1\right)\sqrt{2x^2+3}+2\left(x+1\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2+3}=2\left(x+1\right)\\2\sqrt{2x^2+3}=x+1\end{matrix}\right.\) (\(x\ge-1\))
\(\Rightarrow\left[{}\begin{matrix}2x^2+3=4\left(x+1\right)^2\\4\left(2x^2+3\right)=\left(x+1\right)^2\end{matrix}\right.\) (\(x\ge-1\))
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
a.
\(\Leftrightarrow\sqrt[3]{3x-5}=\left(2x-3\right)^3+2x-3-\left(3x-5\right)\)
Đặt \(\left\{{}\begin{matrix}2x-3=a\\\sqrt[3]{3x-5}=b\end{matrix}\right.\)
\(\Rightarrow b=a^3+a-b^3\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt[3]{3x-5}=2x-3\)
\(\Leftrightarrow3x-5=\left(2x-3\right)^3\)
\(\Leftrightarrow8x^3-36x^2+51x-22=0\)
\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
b.
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+3x-2-\sqrt[3]{81x-8}=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{\left(3x-2\right)^3-\left(81x-8\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{27\left(x^3-2x^2-\dfrac{5}{3}x\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)
\(\Leftrightarrow\left(x^3-2x^2-\dfrac{5}{3}x\right)\left(1+\dfrac{27}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}\right)=0\)
\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x=0\)
23 tháng 7 2021
a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)
⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)
Phương trình vô nghiệm
b, x = \(\dfrac{8}{125}\)
PL
9 tháng 4 2023
a: Khi m = -4 thì:
\(x^2-5x+\left(-4\right)-2=0\)
\(\Leftrightarrow x^2-5x-6=0\)
\(\Delta=\left(-5\right)^2-5\cdot1\cdot\left(-6\right)=49\Rightarrow\sqrt{\Delta}=\sqrt{49}=7>0\)
Pt có 2 nghiệm phân biệt:
\(x_1=\dfrac{5+7}{2}=6;x_2=\dfrac{5-7}{2}=-1\)
NQ
giải pt
a)\(\dfrac{1}{x+1}+\dfrac{3}{2x+1}=\dfrac{8}{x-2}\)
b)\(\sqrt{2x+1}+\sqrt{3-x}=\sqrt{3x+5}\)
0
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