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xin do

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d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a: Ta có: \(x\left(x-3\right)-x^2+5=0\)

\(\Leftrightarrow-3x+5=0\)

hay \(x=\dfrac{5}{3}\)

b: Ta có: \(x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

1) 14x-8x=10+5

x(14-8)=15

x6=15

x=15/6

2)5x-3x=30-15

2x=15

x=15/2

3)làm tương tự

1) x=2,5

2) x=7,5

3) x=4

4) x=7/3

5) x=8,25

\(x^3-6x^2-x+30=0\)

\(\Leftrightarrow x^3-5x^2-x^2+5x-6x+30=0\)

\(\Leftrightarrow x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^2-x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=5\end{matrix}\right.\)

Vậy...

\(x^3-6x^2-x+30=0\)

\(x^3+2x^2-8x^2-16x+15x+30=0\)

\(\left(x^2+2x^2\right)-\left(8x^2+16x\right)+\left(15x+30\right)=0\)

\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

\(\left(x+2\right)\left(x^2-8x+15\right)=0\)

TH1: \(x+2=0\Leftrightarrow x=-2\) (1)

TH2: \(x^2-8x+15=0\)

\(x^2-8x=-15\)

\(x^2-2x.4+16=-15+16\)

\(\left(x-4\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x\in\left\{-2;5;3\right\}\)

a,x2+6x-7=0

=>x2+7x-x-7=0

=>(x^2+7x)-(x+7)=0

=>x(x+7)-(x+7)=0 =>(x+7)(x-1)=0

=>\(\orbr{\begin{cases}x+7=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=1\end{cases}}}\)

b, x^3-2x^2-5x+6=0

=>x(x^2-2x-5+6)=0

=>x(x^2-2x+1)=0\(^{\orbr{\begin{cases}x=0\\\left(x-1^2\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

c, 2x^2-5x+3=0

=>2x^2-2x-3x+3=0

\(x^3-19x-30=0\)

\(\Rightarrow x^3+5x^2+6x-5x^2-25x-30=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+2x+3x+6\right)=0\)

\(\Rightarrow\left(x-5\right)[x\left(x+2\right)+3\left(x+2\right)]=0\)

\(\Rightarrow\left(x-5\right)\left(x+3\right)\left(x+2\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-5=0\\x+3=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=-3\\x=-2\end{cases}}\)

6x³-11x²-19x-6 =0 co ai biet lam ko