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giải giúp mình hệ phương trình 3 ẩn này với mình cảm ơn nhiều
3=(a+b)(a+c)
4=(a+b)(b+c)
5=(a+c)(b+c)
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ĐKXĐ: \(x\ge1\)
\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...
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Pt có 2 nghiệm khi: \(\left\{{}\begin{matrix}m\ne0\\\Delta'=9\left(m-1\right)^2-9m\left(m-3\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ge-1\end{matrix}\right.\)
Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{6\left(m-1\right)}{m}\\x_1x_2=\dfrac{9\left(m-3\right)}{m}\end{matrix}\right.\)
\(x_1+x_2=x_1x_2\Rightarrow\dfrac{6\left(m-1\right)}{m}=\dfrac{9\left(m-3\right)}{m}\)
\(\Rightarrow6\left(m-1\right)=9\left(m-3\right)\)
\(\Rightarrow m=7\)
A đúng
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`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`
`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`
`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`
`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`
`<=> cos2x - cos4x - cos2x = 1/2`
`<=> cos4x = cos(2π)/3`
`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)
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a: Để phương trình có nghiệm duy nhất thì \(\left(m-3\right)\left(m+2\right)< >0\)
hay \(m\notin\left\{3;-2\right\}\)
Để phương trình vô nghiệm thì \(\left\{{}\begin{matrix}\left(m-3\right)\left(m+2\right)=0\\\left(m-3\right)\left(m-1\right)< >0\end{matrix}\right.\Leftrightarrow m=-2\)
Để phương trình có vô số nghiệm thì m=3
\(9,PT\Leftrightarrow x-6=3x-7\left(x\ge6\right)\\ \Leftrightarrow x=\dfrac{1}{2}\left(ktm\right)\\ \Leftrightarrow x\in\varnothing\\ 10,PT\Leftrightarrow3x-2=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\\ \Leftrightarrow4x^2-7x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\left(ktm\right)\Leftrightarrow x\in\varnothing\\ 11,PT\Leftrightarrow\sqrt{x^2+x-1}=2-x\left(x\le2\right)\\ \Leftrightarrow x^2+x-1=x^2-4x+4\\ \Leftrightarrow5x=5\Leftrightarrow x=1\left(tm\right)\\ 12,PT\Leftrightarrow\left(\sqrt{20-x}-4\right)+\left(\sqrt{x+5}-3\right)=0\left(5\le x\le20\right)\\ \Leftrightarrow\dfrac{4-x}{\sqrt{20-x}+4}+\dfrac{x-4}{\sqrt{x+5}+3}=0\\ \Leftrightarrow\left(x-4\right)\left(\dfrac{1}{\sqrt{x+5}+3}-\dfrac{1}{\sqrt{20-x}+4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\\dfrac{1}{\sqrt{x+5}+3}=\dfrac{1}{\sqrt{20-x}+4}\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\sqrt{x+5}+3=\sqrt{20-x}+4\\ \Leftrightarrow\left(\sqrt{x+5}-4\right)-\left(\sqrt{20-x}-3\right)=0\\ \Leftrightarrow\dfrac{x-11}{\sqrt{x+5}+4}+\dfrac{x-11}{\sqrt{20-x}+3}=0\\ \Leftrightarrow\left(x-11\right)\left(\dfrac{1}{\sqrt{x+5}+4}+\dfrac{1}{\sqrt{20-x}+3}\right)=0\\ \Leftrightarrow x=11\left(\dfrac{1}{\sqrt{x+5}+4}+\dfrac{1}{\sqrt{20-x}+3}>0\right)\\ \text{Vậy PT có nghiệm }x\in\left\{4;11\right\}\)
\(13,PT\Leftrightarrow\sqrt{x-1}+\sqrt{3x-2}=\sqrt{5x+1}\left(x\ge-\dfrac{1}{5}\right)\\ \Leftrightarrow4x-3+2\sqrt{\left(x-1\right)\left(3x-2\right)}=5x+1\\ \Leftrightarrow x+4=2\sqrt{3x^2-5x+2}\\ \Leftrightarrow x^2+8x+16=12x^2-20x+8\\ \Leftrightarrow11x^2-28x-8=0\\ \Delta'=14^2+8\cdot11=284\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14-2\sqrt{71}}{11}\\x=\dfrac{14+2\sqrt{71}}{11}\end{matrix}\right.\)
\(14,ĐK:x\ge-1\)
Đặt \(\sqrt{x+1}=a\ge0\)
\(PT\Leftrightarrow2\sqrt{a^2-1+2a}-a=4\\ \Leftrightarrow2\sqrt{a^2+2a-1}=a+4\\ \Leftrightarrow4a^2+8a-4=a^2+8a+16\\ \Leftrightarrow3a^2-20=0\\ \Leftrightarrow a^2=\dfrac{20}{3}\Leftrightarrow x+1=\dfrac{20}{3}\Leftrightarrow x=\dfrac{17}{3}\left(tm\right)\)
\(15,ĐK:-3\le x\le6\)
Đặt \(\sqrt{x+3}+\sqrt{6-x}=a\ge0\)
\(\Leftrightarrow\dfrac{a^2-9}{2}=\sqrt{\left(x+3\right)\left(6-x\right)}\\ PT\Leftrightarrow a-\dfrac{a^2-9}{2}=3\\ \Leftrightarrow2a-a^2+9=6\\ \Leftrightarrow a^2-2a-3=0\\ \Leftrightarrow a=3\left(a\ge0\right)\\ \Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\\ \Leftrightarrow\sqrt{x+3}-3+\sqrt{6-x}=0\\ \Leftrightarrow\dfrac{x-6}{\sqrt{x+3}+3}-\dfrac{x-6}{\sqrt{6-x}}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\\dfrac{1}{\sqrt{x+3}+3}=\dfrac{1}{\sqrt{6-x}}\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\sqrt{x+3}+3=\sqrt{6-x}\\ \Leftrightarrow\sqrt{x+3}-\left(\sqrt{6-x}-3\right)=0\\ \Leftrightarrow\dfrac{x+3}{\sqrt{x+3}}+\dfrac{x+3}{\sqrt{6-x}+3}=0\\ \Leftrightarrow x=-3\left(\dfrac{1}{\sqrt{x+3}}+\dfrac{1}{\sqrt{6-x}+3}>0\right)\\ \text{Vậy PT có nghiệm }x\in\left\{6;-3\right\}\)