2x2+ x + 3=3x\(\sqrt{ }\)
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10 tháng 12 2021
\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)
Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)
Do đó \(x\in\left\{1;2\right\}\)
10 tháng 12 2021
\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)
Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)
Vậy PT có nghiệm \(x=4\)
NA
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(2x^2-2x+\left(x+1-\sqrt{3x+1}\right)+2\left(x+2-\sqrt[3]{19x+8}\right)=0\)
\(\Leftrightarrow2x^2-2x+\dfrac{x^2-x}{x+1+\sqrt[]{3x+1}}+\dfrac{\left(x+7\right)\left(x^2-x\right)}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{19x+8}+\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\left(x^2-x\right)\left(2+\dfrac{1}{x+1+\sqrt[]{3x+1}}+\dfrac{x+7}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{19x+8}+\sqrt[3]{\left(19x+8\right)^2}}\right)=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
MD
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 3 2021
a.
\(\Leftrightarrow2x^2\ge3\Leftrightarrow x^2\ge\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)
b.
\(\Leftrightarrow\left(1-x\right)\left(x-3\right)\ge0\Rightarrow1\le x\le3\)
c.
\(\Leftrightarrow\sqrt{1-3x}\le2-x\Leftrightarrow\left\{{}\begin{matrix}1-3x\ge0\\2-x\ge0\\1-3x\le x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\le2\\x^2-x+3\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2022
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(2x^2+4x+3=3\sqrt{\left(x^2+x+1\right)\left(2x+1\right)}\)
\(\Leftrightarrow2\left(x^2+x+1\right)+\left(2x+1\right)-3\sqrt{\left(x^2+x+1\right)\left(2x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{2x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=\sqrt{2x+1}\\2\sqrt{x^2+x+1}=\sqrt{2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=2x+1\\4\left(x^2+x+1\right)=2x+1\end{matrix}\right.\)
\(\Leftrightarrow...\)
HP
17 tháng 12 2020
ĐK: \(x\ge\dfrac{1}{3}\)
\(2x^2+3x-4=\left(4x-3\right)\sqrt{3x-1}\)
\(\Leftrightarrow16x^2+24x-32=8\left(4x-3\right)\sqrt{3x-1}\)
\(\Leftrightarrow\left(4x-3\right)^2+16\left(3x-1\right)-8\left(4x-3\right)\sqrt{3x-1}=25\)
\(\Leftrightarrow\left(4x-3-4\sqrt{3x-1}\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3-4\sqrt{3x-1}=5\\4x-3-4\sqrt{3x-1}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=x-2\\2\sqrt{3x-1}=2x+1\end{matrix}\right.\)
TH1: \(\sqrt{3x-1}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1=\left(x-2\right)^2\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(2\sqrt{3x-1}=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(3x-1\right)=\left(2x+1\right)^2\\2x+1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-8x+5\\x\ge-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\) vô nghiệm
Vậy \(x=6\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
ĐKXĐ: \(x\ge\dfrac{1}{3}\)
Đặt \(\sqrt{3x-1}=t\ge0\Rightarrow3x-1=t^2\)
\(\Rightarrow\left\{{}\begin{matrix}2x^2+3x-4=\left(4x-3\right)t\\3x-1=t^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+3x-4=4tx-3t\\2t^2=6x-2\end{matrix}\right.\)
\(\Leftrightarrow2x^2+2t^2+3x-4=4tx-3t+6x-2\)
\(\Leftrightarrow2\left(x-t\right)^2-3\left(x-t\right)-2=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2020
Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-12x+21}=a>0\\\sqrt{5x^2-20x+24}=b>0\end{matrix}\right.\)
\(\Rightarrow a+b=a^2-b^2\)
\(\Leftrightarrow a+b=\left(a+b\right)\left(a-b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b-1\right)=0\)
\(\Leftrightarrow a-b-1=0\)
\(\Leftrightarrow\sqrt{5x^2-20x+24}+1=\sqrt{3x^2-12x+21}\)
\(\Leftrightarrow5x^2-20x+25+2\sqrt{5x^2-20x+24}=3x^2-12x+1\)
\(\Leftrightarrow2\sqrt{5x^2-20x+24}=-2x^2+8x-4\)
Ta có: \(\left\{{}\begin{matrix}VT=2\sqrt{5x^2-20x+24}=2\sqrt{5\left(x-2\right)^2+4}\ge4\\VP=-2x^2+8x-4=4-2\left(x-2\right)^2\le4\end{matrix}\right.\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi \(x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
T
27 tháng 10 2023
a, Sửa đề:
\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)
b,
\(x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\)
c,
\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)
d,
\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)