giúp cần gấp tối nay, xong trước 7h tối
1)Gpt: 2x3 + x + 3 =0
2)Gpt: x3 + x2 - x\(\sqrt{2}\) - 2\(\sqrt{2}=0\)
3)Gpt: 23 -9x + 2 = 0
4)Gpt: x3 - 42 + 7x - 6 = 0
5)Gpt: 2x3 + 7x2 + 7x + 2 = 0
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\(ĐK:x\in R\)
Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)
\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)
ĐKXĐ: \(x\ge\dfrac{2}{7}\)
\(\sqrt{5x^2-5x+3}-\left(x+1\right)+2x-\sqrt{7x-2}+4x^2-7x+2=0\)
\(\Leftrightarrow\dfrac{4x^2-7x+2}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{4x^2-7x+2}{2x+\sqrt{7x-2}}+4x^2-7x+2=0\)
\(\Leftrightarrow\left(4x^2-7x+2\right)\left(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1\right)=0\)
Ta có \(\dfrac{1}{\sqrt{5x^2-5x+3}+\left(x+1\right)^2}+\dfrac{1}{2x+\sqrt{7x-2}}+1>0\)
\(\Rightarrow4x^2-7x+2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7-\sqrt{17}}{8}\\x=\dfrac{7+\sqrt{17}}{8}\end{matrix}\right.\)
\(\)
pt<=>căn((x-1/2)^2+75/4)+căn(2(x-1/2)^2+3(x+2)^2)+căn((x-1/2)^2+3(2x+3/2)^2)>=3*căn3(x+2)
dấu = xãy ra khi x=1/2
\(2x^4+7x^3+x^2-7x-3=0\)
\(\Leftrightarrow2x^4+7x^3+3x^2-2x^2-7x-3=0\)
\(\Leftrightarrow\left(2x^4+7x^3+3x^2\right)-\left(2x^2+7x+3\right)=0\)
\(\Leftrightarrow x^2\left(2x^2+7x+3\right)-\left(2x^2+7x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2+7x+3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(2x+1\right)\)
\(\Leftrightarrow x\in\left\{\pm1;\frac{-1}{2};-3\right\}\)
\(\Leftrightarrow\frac{3\left(\sqrt{\left(x+5\right)\left(x+2\right)}+1\right)}{\sqrt{x+5}+\sqrt{x+2}}=3\)
đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)
\(\Rightarrow\frac{ab+1}{a+b}=1\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
thay vào là được
Bạn tự phân tích đa thức thành nhân tử nhé!
\(1.\)
\(2x^3+x+3=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(2x^2-2x+3\right)=0\) \(\left(1\right)\)
Vì \(2x^2-2x+3=2\left(x^2-x+1\right)+1=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>0\) với mọi \(x\in R\)
nên từ \(\left(1\right)\) \(\Rightarrow\) \(x+1=0\) \(\Leftrightarrow\) \(x=-1\)
1)2x^3+x+3=0=>