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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)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
\(6x^2-\left(2x+5\right)\left(3x-2\right)=7.\)
\(6x^2-\left(6x^2-4x+15x-10\right)=7\)
\(6x^2-6x^2+4x-15x+10=7\)
\(-11x+10=7\)
\(-11x=-3\)
\(x=\frac{-3}{-11}=\frac{3}{11}\)
\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)
\(6x^2-\left(6x^2-4x+15x-10\right)=7\)
+ 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 !!!
a)\(\Leftrightarrow\left(9x^2-30x+25\right)-\left(9x^2+6x+1\right)\)
\(\Leftrightarrow9x^2-30x+25-9x^2-6x-1=8\)
\(\Leftrightarrow9x^2-30x-9x^2-6x=8-25+1\)
\(\Leftrightarrow-36x=-16\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy \(x=\frac{4}{9}\)
b)\(\Leftrightarrow16x^2-6x-\left(16x^2-24x+9\right)=27\)
\(\Leftrightarrow16x^2-6x-16x^2+24x-9=27\)
\(\Leftrightarrow16x^2-6x-16x^2+24x=27+9\)
\(\Leftrightarrow18x=36\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
Chúc bạn học tốt.
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
4x(3x - 5) + 2x(8 - 6x) = 24
12x2 - 20x + 16x - 12x2 = 24
-4x = 24
x = -6
* Chúc bạn học tốt .
4x(3x -5)+2x(8-6x) =24
=>2x(3x-5)+x(8-6x)=12
=>6x2-10x+8x-6x2=12
=>-2x=12
=>x=-6
=>x+6=0
=>x=-6