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Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)
=>x^2+6x=0
=>x(x+6)=0
=>x=0 hoặc x=-6
\(6x^2-\left(2x+5\right)\left(3x-2\right)=7.\)
\(6x^2-\left(6x^2-4x+15x-10\right)=7\)
\(6x^2-6x^2+4x-15x+10=7\)
\(-11x+10=7\)
\(-11x=-3\)
\(x=\frac{-3}{-11}=\frac{3}{11}\)
\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)
\(6x^2-\left(6x^2-4x+15x-10\right)=7\)
+ Áp dụng quy tắc bở ngoặc ta có :
\(6x^2-6x^2+4x-15x+10=7\)
\(-11x+10=7\)
\(-11x=7-10\)
\(-11x=-3\)
\(x=-3:-11\)
\(x=\frac{-3}{-11}\)
\(\Rightarrow x=\frac{3}{11}\)
Vậy \(x=\frac{3}{11}\)
Chúc bạn học tốt !!!
\(x^2\left(x-2\right)+x\left(-3+2x\right)-5x=7\)
\(x^3-2x^2-3x+2x^2-5x=7\)
\(x^3-8x-7=0\)
<=>x^3-6x^2+12x-8-x^3+6x^2=7
<=>12x=15
<=> x=5/4
chúc bạn hk tốt