Trả lời

1, \(3\left(x-1\right)^2-5x\left(x-6\right)=3\)

\(\Leftrightarrow3\left(x^2-2x+1\right)-5x^2+30x=3\)

\(\Leftrightarrow3x^2-6x+3-5x^2+30x=3\)

\(\Leftrightarrow-2x^2+24x+3=3\)

\(\Leftrightarrow-2x^2+24x=0\)

\(\Leftrightarrow-2x\left(x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-2x=0\\x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=12\end{cases}}}\)

Vậy x = 0; x = 12

2, \(\left(x-5\right)^2-36=0\)

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

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

\(\Leftrightarrow\left(x-11\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-11=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}}\)

Vậy x = 11; x = - 1

3, \(\left(x-1\right)^2-4x+3=0\)

\(\Leftrightarrow x^2-2x+1-4x+3=0\)

\(\Leftrightarrow x^2-6x+4=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{5}\\x-3=-\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}+3\\x=-\sqrt{5}+3\end{cases}}\)

Vậy \(x=\sqrt{5}+3;x=-\sqrt{5}+3\)

4, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow2x+255=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=-\frac{255}{2}\)

Vậy x = - 255/2

5, \(\left(x+2\right)^2-x^2+8=0\)

\(\Leftrightarrow x^2+4x+4-x^2+8=0\)

\(\Leftrightarrow4x+12=0\)

\(\Leftrightarrow4x=-12\)

\(\Leftrightarrow x=-3\)

Vậy x = - 3

\(A=x-15x+75x-125\)

\(=61x-125\)

\(=61.35-125=2010\)

\(3\left(x-1\right)^2-5x\left(x-6\right)=3\)

\(\Leftrightarrow3x^2-6x+3-5x^2+30x=3\)

\(\Leftrightarrow-2x^2+24x=0\Leftrightarrow2x\left(x-12\right)=0\)

\(\Leftrightarrow x=0,x=12\)

\(\left(x+4\right)\left(x-6\right)+\left(x-7\right)^2\)

\(=x^2-2x-24+x^2-12x+49\)

\(=2x^2-14x+25\)

a) \(2x+2y=2\left(x+y\right)\)

b) \(4ab-10a=2a\left(2b-5\right)\)

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

Ta dễ có bất đẳng thức \(a^2+b^2+c^2\ge ab+bc+ca\)

Dấu bằng xảy ra khi a = b = c

Mà a+b+c=2021 nên a=b=c=2021/3

VBJHNDXB 

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