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Ta có: \(2x^4+11x^3+11x^2-24x-36\)
\(=2x^4+4x^3+7x^3+14x^2-3x^2-6x-18x-36\)
\(=\left(x+2\right)\left(2x^3+7x^2-3x-18\right)\)
\(=\left(x+2\right)\left(2x^3+4x^2+3x^2+6x-9x-18\right)\)
\(=\left(x+2\right)^2\left(2x^2+3x-9\right)\)
\(=\left(x+2\right)^2\cdot\left(2x^2+6x-3x-9\right)\)
\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)\)
Ta có: \(x^5+7x^4+21x^3+47x^2+80x+60\)
\(=x^5+4x^4+4x^3+3x^4+12x^3+12x^2+5x^3+20x^2+20x+15x^2+60x+60\)
\(=\left(x^2+4x+4\right)\left(x^3+3x^2+5x+15\right)\)
\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)\)
Ta có: \(\frac{2x^4+11x^3+11x^2-24x-36}{x^5+7x^4+21x^3+47x^2+80x+60}\)
\(=\frac{\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)}{\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)}=\frac{2x-3}{x^2+5}\)
\(2x^2-7x+3=0\Leftrightarrow2x^2-x-6x+3=0\Leftrightarrow x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)
\(2x^2-7x+3=0\Leftrightarrow x=\frac{-1\pm\sqrt{119}t}{12}\)
hoặc bn cho là vô nghiệm cx đc
\(16x^2+24x+9=0\Leftrightarrow\left(4x+3\right)^2=0\Leftrightarrow4x+3=0\Leftrightarrow x=-\frac{3}{4}\)
\(\Leftrightarrow x^4-x^3+x^3-x^2-8x^2+8x+16x-16=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+4x^2-3x^2-12x+4x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x^2-3x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\\\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\left(vô.n_o\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)
\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)
\(x^4-8x^3+21x^2-24x+9=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))
\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
Phương trình 16x2 − 24x + 9 = 0
có a = 16; b’ = −12; c = 9 suy ra
Δ ' = b ' 2 − a c = (−12)2 – 9.16 = 0
Nên phương trình có nghiệm kép
Đáp án cần chọn là: C
đặt \(a=1-\sqrt{2}\),ta có
\(1-a=\sqrt{2}\)\(\Rightarrow\left(1-a\right)^2=2\)
\(\Rightarrow a^2-2a+1=2\Rightarrow a^2-2a-1=0\)
\(\Rightarrow x^2-2x-1=0\)nhận \(1-\sqrt{2}\)là nghiệm
\(\Rightarrow b=-2;c=-1\)
\(x=-0,206885594\) nhé =)