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

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

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

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

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

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

bạn vào loigiaihay rồi chọn toán lớp 8 rồi chọn đẳng thức đáng nhớ

dễ mà áp dụng hết hằng đẳng thức nếu bạn thuộc hằng đẳng thức mik chỉ làm mỗi bài 1 ý nha xong dựa vô mà làm

\(1a.\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2\)

                                   \(=4y^2+12xy+9y^2\)

\(2a.x^2-6x+9\)

\(=x^2-2.x.3+3^2\)

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

Bạn tự vẽ hình ra nhé

Ta có góc A+ góc B+ góc C+ góc D =360  độ

=> góc A+ góc B = 360-80-70=210  độ 

Có góc BAy =1/2 góc A

góc ABy=1/2 góc B

nên góc BAy+ góc ABy=1/2.( góc A+ góc B )=1/2.210=105 độ

Tam giác ABy có  góc ABy+ góc BAy+ góc AyB =180 độ

=> góc AyB=180 độ - ( góc ABy+góc BAy ) =180 độ -105 độ=75 độ

Vậy góc AyB= 75 độ 

b,(x-5)² -x*(x-4)=9

(x²-2x5+5²)-x²+4x=9

x²-10x+25-x²+4x=9

-6x+25=9

x=8/3

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

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

\(\Rightarrow x+1=-1\)

\(\Rightarrow x=-2\)

đề 

\(x^3+3x^2.1+3x.1^2+1^3=-1\)

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

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

\(->x+1=-1\)

          \(x=-2\)

\(-\left(5x^2-x+15\right)=-\frac{1}{20}\left(100x^2-20x+300\right)=-\frac{1}{20}[\left(10x-1\right)^2-\frac{299}{20}< 0\forall x\)

\(-5x^2+x-15\)

\(=-5\left(x^2-\frac{1}{5}x+3\right)\)

\(=-5\left(x^2-2x\cdot\frac{1}{10}+\frac{1}{100}+\frac{299}{100}\right)\)

\(=-5\left[\left(x-\frac{1}{10}\right)^2+\frac{299}{100}\right]\)

Vì \(\left(x-\frac{1}{10}^2\right)+\frac{299}{100}>0\)\(\forall x\)

\(\Rightarrow-5\left[\left(x-\frac{1}{10}\right)^2+\frac{299}{100}\right]< 0\)\(\forall x\)

\(\Rightarrow-5x^2+x-15\)luôn âm