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\(1,\Delta=\left(-11\right)^2-4\cdot30=1\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11-1}{2}=5\\x=\dfrac{11+1}{2}=6\end{matrix}\right.\\ 2,\Delta=\left(-1\right)^2-4\left(-20\right)=81\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{81}}{2}=-4\\x=\dfrac{1+\sqrt{81}}{2}=5\end{matrix}\right.\\ 3,\Delta=14^2-4\cdot24=100\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14-\sqrt{100}}{2}=-12\\x=\dfrac{-14+\sqrt{100}}{2}=-2\end{matrix}\right.\\ 4,\Delta=8^2-4\left(-2\right)3=88\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-8-\sqrt{88}}{6}=\dfrac{-4+\sqrt{22}}{3}\\x=\dfrac{-8+\sqrt{88}}{6}=\dfrac{-4-\sqrt{22}}{3}\end{matrix}\right.\)
= 3-x +4can 3-x +4 +x =13
4căn 3-x = 6
16(3-x) = 36
48-36 = 16x
x = 16/12 = 4/3
ĐKXĐ: x<>1
Ta có: \(\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}=2\)
=>\(\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2\left(x-1\right)^2}{\left(x-1\right)^3}=2\)
=>\(x^3+3x^2\left(x-1\right)^2=2\left(x-1\right)^3\)
=>\(x^2\left\lbrack x+3\left(x-1\right)^2\right\rbrack=2\left(x-1\right)^3\)
=>\(x^2\left(3x^2-6x+3+x\right)=2\left(x^3-3x^2+3x-1\right)\)
=>\(3x^4-5x^3+3x^2-2x^3+6x^2-6x+2=0\)
=>\(3x^4-7x^3+9x^2-6x+2=0\)
=>\(\left(x^2-x+1\right)\left(3x^2-4x+2\right)=0\)
mà \(x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\forall x\)
nên \(3x^2-4x+2=0\)
=>\(x^2-\frac43x+\frac23=0\)
=>\(x^2-2\cdot x\cdot\frac23+\frac49+\frac29=0\)
=>\(\left(x-\frac23\right)^2+\frac29=0\) (vô lý)
=>x∈∅