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a,x4-10x2+9=0
=>(x-1)(x3+x2-9x-9)=0
=> (x-1)(x+1)(x-3)(x+3)=0
=>\(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)hoặc\(\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm1\\x=\pm3\end{cases}}\)
Vậy tập nghiệm cuả pt là S={\(\pm1,\pm3\)}
Câu 1/
x4 + (x - 1)(x2 - 2x + 2) = 0
\(\Leftrightarrow\)x4 + x3 - 3x2 + 4x - 2 = 0
\(\Leftrightarrow\)(x4 - x3 + x2) + (2x3 - 2x2 + 2x) + (- 2x2 + 2x + 2) = 0
\(\Leftrightarrow\)(x2 - x + 1)(x2 + 2x - 2) = 0
Tới đây tự làm tiếp nhé.
Câu 2/ Đặt \(\hept{\begin{cases}\frac{x+1}{x-2}=a\\\frac{x-2}{x-4}=b\end{cases}}\)
Thì ta có pt
\(\Leftrightarrow\)a2 + ab - 12b2 = 0
\(\Leftrightarrow\)(a2 - 3ab) + (4ab - 12b2) = 0
\(\Leftrightarrow\)(a - 3b)(a + 4b) = 0
Tự làm phần còn lại nhé.
\(pt\Leftrightarrow\left(\frac{4}{x^2}+\frac{x^2}{4-x^2}\right)+\frac{5}{2}\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)+2=0\)
\(\Leftrightarrow\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)^2-1+\frac{5}{2}\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)+2=0\)
Đặt \(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}=t\)pt thành
\(t^2-1+\frac{5}{2}t+2=0\)\(\Rightarrow\orbr{\begin{cases}t=-2\\t=-\frac{1}{2}\end{cases}}\)(loại)
-->PT vô nghiệm
\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)
\(=-1+\sqrt{100}\)
\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)
\(a^2-\left(b+1\right)a+b=0\Leftrightarrow\left(a-1\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)
\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)
Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\)
\(Pt\Leftrightarrow10a^2+b^2-11ab=0\\ \Leftrightarrow10a^2-10ab-ab+b^2=0\\ \Leftrightarrow10a\left(a-b\right)-b\left(a-b\right)=0\\ \Leftrightarrow\left(10a-b\right)\left(a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10a-b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10a=b\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{10\left(x-2\right)}{x+1}=\frac{x+2}{x-1}\left(1\right)\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow10\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow10\left(x^2-3x+2\right)=x^2+3x+2\\ \Leftrightarrow9x^2-33x+18=0\\ \Leftrightarrow9x^2-27x-6x+18=0\\ \Leftrightarrow9x\left(x-3\right)-6\left(x-3\right)=0\\ \Leftrightarrow\left(9x-6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\9x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{2}{3}\end{matrix}\right.\left(Tm\right)\)
\(\left(2\right)\Leftrightarrow\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow x^2-3x+2=x^2+3x+2=0\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\left(Tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{0;3;\frac{2}{3}\right\}\)
\(\text{b) }\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\\ DKXD:x\ne2;x\ne4\\ \Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-2}\cdot\frac{x-2}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)
Đặt \(\frac{x+1}{x-2}=a;\frac{x-2}{x-4}=b\)
\(Pt\Leftrightarrow a^2+ab-12b^2=0\\ \Leftrightarrow a^2+4ab-3ab-12b^2=0\\ \Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\\ \Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=-4b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-2}=\frac{3\left(x-2\right)}{x-4}\left(1\right)\\\frac{x+1}{x-2}=\frac{-4\left(x-2\right)}{x-4}\left(2\right)\end{matrix}\right.\)
Tự giải tiếp nha.
\(\left(x^2+\frac{4}{x^2}\right)-4.\left(x-\frac{2}{x}\right)+9=0\)
Đặt \(x-\frac{2}{x}=t\) \(\Rightarrow x^2+\frac{4}{x^2}=t^2+4\)
Phương trình đã cho trở thành:
\(t^2+4-4t+9=0\)
\(\Leftrightarrow t^2-4t+13=0\)
\(\Delta=\left(-4\right)^2-4.1.13\)
\(=16-52=-36< 0\)
\(\Rightarrow\)Phương trình vô nghiệm