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ĐK: \(-1\le x\le1\)
Đặt \(\sqrt{1-x}=a;\sqrt{x+1}=b\Rightarrow3-x=2a^2+b^2\)
\(pt\Leftrightarrow2a-b+3ab=2a^2+b^2\)
\(\Leftrightarrow2a^2+b^2-2a+b-3ab=0\)
\(\Leftrightarrow2a^2-a\left(3b+2\right)+b^2+b=0\)
\(\Delta=\left(3b+2\right)^2-4.2.\left(b^2+b\right)=9b^2+12b+4-8b^2-8b\)
\(=b^2+4b+4=\left(b+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{3b+2-\left(b+2\right)}{4}=\dfrac{2b}{4}=\dfrac{b}{2}\Leftrightarrow2a=b\left(1\right)\\a=\dfrac{3b+2+b+2}{4}=\dfrac{4b+4}{4}=b+1\left(2\right)\end{matrix}\right.\)
pt (1) \(\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\)
\(\Leftrightarrow4\left(1-x\right)=x+1\)
\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)
\(pt\left(2\right)\Leftrightarrow\sqrt{1-x}=1+\sqrt{x+1}\)
\(\Leftrightarrow1-x=1+x+1+2\sqrt{x+1}\)
\(\Leftrightarrow-1-2x=2\sqrt{x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2+4x+1=4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(l\right)\\x=-\dfrac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy, pt có tập nghiệm là: \(S=\left\{-\dfrac{\sqrt{3}}{2};\dfrac{5}{3}\right\}\)
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`x^2-2x-sqrt3+1=0`
Vì `Delta=1+sqrt3-1>0`
`=>` pt có 2 nghiệm pb
ÁP dụng vi-ét:
`x_1+x_2=2,x_1.x_2=1-sqrt3`
`M=x_1^2x_2^2-2x_1.x_2-x_1-x_2`
`=(x_1.x_2)^2-2(x_1.x_2)-(x_1+x_2)`
`=(sqrt3-1)^2-2(1-sqrt3)-2`
`=4-2sqrt3-2+2sqrt3-2`
`=0`
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\(\text{Δ}=\left(2m-1\right)^2-8\left(m-1\right)\)
\(=4m^2-4m+1-8m+8\)
\(=4m^2-12m+9=\left(2m-3\right)^2\)
Để phương trình có hai nghiệm phân biệt thì 2m-3<>0
hay m<>3/2
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}3x_1-4x_2=11\\x_1+x_2=\dfrac{-2m+1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x_1-4x_2=11\\2x_1+2x_2=-2m+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_1-4x_2=11\\4x_1+4x_2=-4m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x_1=-4m+13\\4x_2=3x_1-11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-4m+13}{7}\\4x_2=\dfrac{-12m+36}{7}-\dfrac{77}{7}=\dfrac{-12m-41}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-4m+13}{7}\\x_2=\dfrac{-12m-41}{28}\end{matrix}\right.\)
Theo Vi-et, ta được: \(x_1x_2=\dfrac{m-1}{2}\)
\(\Leftrightarrow\dfrac{\left(4m-13\right)\left(12m+41\right)}{196}=\dfrac{m-1}{2}\)
\(\Leftrightarrow\left(4m-13\right)\left(12m+1\right)=98\left(m-1\right)\)
\(\Leftrightarrow48m^2+4m-156m-13-98m+98=0\)
\(\Leftrightarrow48m^2-250+85=0\)
Đến đây bạn chỉ cần giải pt bậc hai là xong rồi
\(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+10\)
\(=\left(2m-3\right)^2+1>0\)
Vậy pt có 2 nghiệm pb
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1-2m}{2}\left(1\right)\\x_1x_2=\dfrac{m-1}{2}\left(2\right)\end{matrix}\right.\)
Ta có \(3x_1-4x_2=11\left(3\right)\)
Từ (1) ; (3) ta có hệ \(\left\{{}\begin{matrix}4x_1+4x_2=2-4m\\3x_1-4x_2=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x_1=13-4m\\x_2=\dfrac{1-2m}{2}-x_1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{13-4m}{7}\\x_2=\dfrac{1-2m}{2}-\dfrac{13-4m}{7}\end{matrix}\right.\)
\(x_2=\dfrac{7-14m-26+8m}{14}=\dfrac{-19-6m}{14}\)
Thay vào (2) ta được \(\left(\dfrac{13-4m}{7}\right)\left(\dfrac{-19-6m}{14}\right)=\dfrac{m-1}{2}\)
\(\Leftrightarrow m=4,125\)
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Đã có cách giải phương trình bậc 3 bằng biệt thức rồi mà:
\(2x^3+4x^2+x-2=0\)
với a = 2; b = 4; c =1; d = -2. Là các hệ số
\(\Delta=b^2-3ac=4^2-3.2.1=10>0\)
\(k=\frac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^2}}=\frac{4\sqrt{10}}{5}>1\) Em thay số vào nhé
Vì \(\Delta>0;\left|k\right|>1\)nên phương trình bậc 3 có nghiệm duy nhất:
=> \(x=\frac{\sqrt{\Delta}\left|k\right|}{3ak}.\left(\sqrt[3]{\left|k\right|+\sqrt{k^2-1}}+\sqrt[3]{\left|k\right|-\sqrt{k^2-1}}\right)-\frac{b}{3a}\)
\(=\frac{\sqrt{10}}{6}\left(\sqrt[3]{\frac{4\sqrt{10}}{5}+\frac{3\sqrt{15}}{5}}+\sqrt[3]{\frac{4\sqrt{10}}{5}-\frac{3\sqrt{15}}{5}}\right)-\frac{4}{6}\)