a, \(\left(2x^3-x^2+5x\right):x=2x^2-x+5\)

b, \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)=-\frac{3}{2}x^3+x^2-\frac{1}{2}x\)

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

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

8x² - 11x + 3

= 8x² - 8x - 3x + 3

= 8x. ( x - 1 ) - 3. ( x - 1 )

= ( x - 1 ). ( 8x - 3 )

Phân tích đa thức thành nhân tử 

8x²-11x+3

= 8x² - 8x -3x +3

= 8x(x-1)-3(x-1)

=(x-1)(8x-3)

Ta có \(A^2=\left(x-\sqrt{50}+x-\sqrt{50}-2.\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=\left(2x-2.\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=2.\left(x-\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=2.\left(x^2-x^2+50\right)\)

\(=100\)

Ta có \(\sqrt{x-\sqrt{50}}< \sqrt{x+\sqrt{50}}\)

\(\Rightarrow\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}< 0\)

mà \(\sqrt{x+\sqrt{x^2-50}}\ge0\)

Nên \(A\le0\)

Có \(A^2=100\)

Nên A=-10

\(\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\sqrt{x+\sqrt{x^2-50}}\)

\(=\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right).\frac{1}{\sqrt{2}}.\sqrt{2x+2\sqrt{x-\sqrt{50}}.\sqrt{x+\sqrt{50}}}\)

\(=\frac{1}{\sqrt{2}}.\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\sqrt{\left(\sqrt{x-\sqrt{50}}+\sqrt{x+\sqrt{50}}\right)^2}\)

\(=\frac{1}{\sqrt{2}}.\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\left(\sqrt{x-\sqrt{50}}+\sqrt{x+\sqrt{50}}\right)\)

\(=\frac{1}{\sqrt{2}}.\left(x-\sqrt{50}-x-\sqrt{50}\right)=\frac{-2\sqrt{50}}{\sqrt{2}}=-10\)

775=5²+5³+5⁴(1)

625.5=5⁵(2)

Từ 1 và 2 ta có 775.625.5=(5²+5³+5⁴).5⁵

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=15.256.\)

\(2^x+2.2^x+4.2^x+8.2^x=15.256\)

\(15.2^x=15.256\Rightarrow2^x=256=2^8\Rightarrow x=8\)

\(\Rightarrow15.256=2^8+2^9+2^{10}+2^{11}\)

\(x^2-x-20=0\)

\(\Leftrightarrow x^2-5x+4x-20=0\)

\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)=0\)

\(\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)

cn lại lm tg tự nha bn

=.= hok tốt!!

`Answer:`

\(x^2-x-20=0\)

\(\Leftrightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{81}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{81}{4}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\left(\frac{9}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{9}{2}\\x-\frac{1}{2}=-\frac{9}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

\(x^2+80x-20=0\)

\(\Leftrightarrow x^2+2.40x+1600-1620=0\)

\(\Leftrightarrow\left(x+40\right)^2-\sqrt{1620}=0\)

\(\Leftrightarrow\left(x+40\right)^2=18\sqrt{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x+40=18\sqrt{5}\\x+40=-18\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=18\sqrt{5}-40\\x=-18\sqrt{5}-40\end{cases}}\)

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

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

\(\Leftrightarrow x.\left(x-1\right)+6.\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+6=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-6\\x=1\end{cases}}\)

k giùm nha :v

ok mik cay cho

\(1.\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(2.\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\frac{a^2}{4}\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\)
Dấu "=" xảy ra khi \(a=b=c=0\)
\(3.\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\left(\frac{a}{2}-e\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)
Dấu "=" xảy ra khi \(\frac{a}{2}=b=c=d=e\)
4. Ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\)

\(\left(c-d\right)^2\ge0\Rightarrow c^2+d^2\ge2cd\)
\(\Rightarrow a^2+b^2+c^2+d^2\ge2ab+2cd\)

\(\Rightarrow a^2+b^2+c^2+d^2+ab+cd\ge3ab+3cd\)
Ta lại có:\(\left(\sqrt{ab}-\sqrt{cd}\right)^2\ge0\Rightarrow ab+cd\ge2\sqrt{abcd}=2\)

\(\Rightarrow3\left(ab+cd\right)\ge6\)
\(\Rightarrow a^2+b^2+c^2+d^2+ab+cd\ge3\left(ab+cd\right)\ge6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=b\\c=d\\ab=cd\end{cases}}\Leftrightarrow a=b=c=d\)

\(\left|x+2012\right|+\left|x-2014\right|=\left|x+2012\right|+\left|2014-x\right|\)
Ta có: \(\left|x+2012\right|+\left|2014-x\right|\ge\left|x+2012+2014-x\right|\)
\(\Rightarrow\left|x+2012\right|+\left|2014-x\right|\ge4026\ge2016\)
Ta có đpcm