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

\(< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Trả lời:

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}}\)

Vậy x = 5/3; x = 5/2 là nghiệm của pt.

Trả lời:

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)

Vậy x = - 1; x = 1/3 là nghiệm của pt.

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Leftrightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

Vậy \(S=\left\{-1;\frac{1}{3}\right\}\)

\(\left(x-1\right)^2-\left(2x\right)^2=0\)

\(\Rightarrow\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(\Rightarrow\left(-x-1\right)\left(3x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x-1=0\\3x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=1\\3x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

\(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)

\(=\left(x^3-4x^2+3x^2+7x-12x+21\right):\left(x^2-4x+7\right)\)

\(=\left[\left(x^3-4x^2+7x\right)+\left(3x^2-12x+21\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[x\left(x^2-4x+7\right)+3\left(x^2-4x+7\right)\right]:\left(x^2-4x+7\right)\)

\(=\left[\left(x^2-4x+7\right)\left(x+3\right)\right]:\left(x^2-4x+7\right)\)

\(=x+3\)

đầy đủ giúp em nhé

\(5x\left(x-2\right)=x-2\)

\(\Leftrightarrow5x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow5x^2-10x-x+2=0\)

\(\Leftrightarrow\left(5x^2-x\right)-\left(10x-2\right)=0\)

\(\Leftrightarrow x\left(5x-1\right)-2\left(5x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)

Vậy \(S=\left\{\frac{1}{5};2\right\}\)

tìm x biết: 5x(x-2)=x-2

x=1/5,

x=2

nha bạn 

chúc học ngoan 

\(x^2-xy-5x+5y+2=0\)

\(\Leftrightarrow x\left(x-y\right)-5\left(x-y\right)=-2\)

\(\Leftrightarrow\left(x-y\right)\left(x-5\right)=-2\)

Ta có

Vậy \(\left(x;y\right)=\left\{\left(3;2\right);\left(7;8\right)\right\}\)

Trả lời:

\(I=x^4-6x^3+11x^2-12x+20\)

\(=x^4-6x^3+9x^2+2x^2-12x+18+2\)

\(=\left(x^4-6x^3+9x^2\right)+\left(2x^2-12x+18\right)+2\)

\(=\left[\left(x^2\right)^2-2.x^2.3x+\left(3x\right)^2\right]+2\left(x^2-6x+9\right)+2\)

\(=\left(x^2-3x\right)^2+2\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2-3x=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0;x=3\\x=3\end{cases}\Leftrightarrow}\hept{x=3}}\)

Vậy GTNN của I = 2 khi x = 3

\(A=x^4-6x^3+10x^2-6x+9\)

\(=x^4-6x^3+9x^2+x^2-6x+9\)

\(=\left(x^4-6x^3+9x^2\right)+\left(x^2-6x+9\right)\)

\(=\left(x^2-3x\right)^2+\left(x-3\right)^2\ge0\forall x\)

Dấu "=" xảy ra khi x = 3  (giống ý trên)

Vậy GTNN của A = 0 khi x = 3

\(f\left(x\right)=x^3-x^2+3x-3\)

\(=x^2\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x^2+3\right)\left(x-1\right)\)

Để \(f\left(x\right)>0\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)

Mà \(x^2\ge0\forall x\Leftrightarrow x^2+3>0\)

\(\Rightarrow x-1>0\Leftrightarrow x=1\)

\(h\left(x\right)=4x^3-14x^2+6x-21< 0\)

\(\Leftrightarrow0\left(x-\frac{7}{2}\right)\left(4x^2+6\right)< 0\)

Mà \(4x^2+6>0\forall x\Leftrightarrow h\left(x\right)< 0\Leftrightarrow x-\frac{7}{2}< 0\Leftrightarrow x< \frac{7}{2}\)

f(x)=x3−x2+3x−3f(x)=x3−x2+3x−3

=x2(x−1)+3(x−1)=x2(x−1)+3(x−1)

=(x2+3)(x−1)=(x2+3)(x−1)

Để f(x)>0⇔(x2+3)(x−1)>0f(x)>0⇔(x2+3)(x−1)>0

Mà x2≥0∀x⇔x2+3>0x2≥0∀x⇔x2+3>0

⇒x−1>0⇔x=1⇒x−1>0⇔x=1

h(x)=4x3−14x2+6x−21<0h(x)=4x3−14x2+6x−21<0

⇔0(x−72)(4x2+6)<0⇔0(x−72)(4x2+6)<0

Mà 4x2+6>0∀x⇔h(x)<0⇔x−72<0⇔x<72