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Ta có: \(x^4-30x^2+31x-30=0\) \(\Rightarrow x^4+x-30x^2+30x-30=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2+x-30=0\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)
Vậy x=5 hoặc x = -6
\(\Leftrightarrow x^4-5x^3+5x^3-25x^3-5x^3+25x+6x-30=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x^3+6x^2-x^2-6x+x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
hay \(x\in\left\{5;-6\right\}\)
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
Ta có : \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x-2\right)-24=0\)
Đặt t = x2 + 5x - 1
Khi đó : (x2 + 5x) = t + 1 ; (x2 + 5x - 2) = t - 1
Ta có : C = (x2 + 5x - 2)2 (x2 + 5x - 2) - 24 = 0
=> (x2 + 5x - 2)3 = 24
MK chỉ giả được đến đây thôi
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
a ) Do \(x^2+1>0;\left(x^2+1\right)\left(x-3\right)< 0\Rightarrow x-3< 0\Rightarrow x< 3\)
b ) Do \(x^2+1>0\Rightarrow-\left(x^2+1\right)< 0\)
Mà \(\left(-x^2-1\right)\left(x-1\right)\ge0\Rightarrow x-1\le0\Rightarrow x\le1\)