\(D=-20\%:\left(-\dfrac{1}{2}\right)^2-\left(-1\right)^{2021}.2\dfrac{1}{3}-0,2\)

\(\Rightarrow D=-\dfrac{1}{5}:\dfrac{1}{4}+1.\dfrac{7}{3}-\dfrac{1}{5}\)

\(\Rightarrow D=-\dfrac{4}{5}+\dfrac{7}{3}-\dfrac{1}{5}\)

\(\Rightarrow D=-1+\dfrac{7}{3}=\dfrac{4}{3}\)

Trả lời nhanh giúp mik ạ

`|(-2 2/3 )^2 - x | - 1/3 = 0`

`<=> | ( -8/3 )^2 - x | = 1/3`

`<=> | 64/9 - x | = 1/3`

`<=> 64/9 - x = 1/3` hoặc `64/9 - x = -1/3`

`<=> x = 61/9` hoặc `x = 67/9`

Vậy 

Để cho đa thức có nghiệm thì

\(x^3+2x^2-11x-12=0\)

\(\Rightarrow x^3+5x^2-3x^2+4x-15x-12=0\)

\(\Rightarrow x\left(x^2+5x+4\right)-3\left(x^2+5x+4\right)=0\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x-3\right)=0\)

\(\Rightarrow\left[\left(x^2+x\right)+\left(4x+4\right)\right]\left(x-3\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+4\right)\left(x-3\right)=0\)

\(\Rightarrow x=-1\) hoặc \(x=-4\) hoặc \(x=3\)

theo pytago ta có bình phương độ dài đường chéo hình vuông là

169² + 169² = 2.169² (cm)

độ dài đường chéo hình vuông là  \(\sqrt{2.169^2}\)  = 169\(\sqrt{2}\)(cm)

b, theo pytago ta có 

bình phương độ dài đường chéo hình vuông là

196² + 196² = 2.196² (cm)

độ dài đường chéo hình vông là \(\sqrt{2.196^2}\) = 196.\(\sqrt{2}\)(cm)

đs....

bạn ơi pytago là gì hả bạn  ?????

\(\dfrac{2020}{2019}\) > 1 >  \(\dfrac{13}{14}\) (so sánh với 1 )

vậy \(\dfrac{2020}{2019}\) > \(\dfrac{13}{14}\)

\(0,5x+\dfrac{1}{3}-\dfrac{2}{3}\left(x+1\right)=\dfrac{2}{9}\)

\(\dfrac{1}{2}x+\dfrac{1}{3}-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{2}{9}\)

\(-\dfrac{1}{6}x-\dfrac{1}{3}=\dfrac{2}{9}\)

\(-\dfrac{1}{6}x=\dfrac{2}{9}+\dfrac{1}{3}\)

\(-\dfrac{1}{6}x=\dfrac{5}{9}\)

\(x=-\dfrac{10}{3}\)

- \(\dfrac{6}{7}\) ( x + \(\dfrac{1}{2}\) ) > 0 ⇔ (x + 1/2 ) < 0 ⇔ x < -1/2 

\(x+\dfrac{1}{3}-\dfrac{2}{3}\left(x+1\right)=\dfrac{2}{9}\)

\(x+\dfrac{1}{3}-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{2}{9}\)

\(\dfrac{1}{3}x-\dfrac{1}{3}=\dfrac{2}{9}\)

\(\dfrac{1}{3}x=\dfrac{2}{9}+\dfrac{1}{3}\)

\(\dfrac{1}{3}x=\dfrac{5}{9}\)

\(x=\dfrac{5}{9}\cdot3\)

\(x=\dfrac{15}{9}=\dfrac{5}{3}\)

cảm ơn bạn

1.\(\left(-3x+1\right)\times\dfrac{1}{2}=\left(-\dfrac{1}{2}\right)^2\Rightarrow\left(-3x+1\right)=\dfrac{1}{2}\)

\(\Rightarrow\left(-3\right)x=\dfrac{1}{2}-1\Rightarrow x=-\dfrac{1}{2}:\left(-3\right)=\dfrac{1}{6}\)

2.\(x+\dfrac{5}{4}=\dfrac{16}{\left(x+5\right)^2}\Rightarrow x\left(x+5\right)^2+\dfrac{5\left(x+5\right)^2}{4}=16\)

\(4x\left(x+5\right)^2+5\left(x+5\right)^2=64\Rightarrow4x\left(x^2+10x+25\right)+5\left(x^2+10x+25\right)-64=0\)

\(4x^3+40x^2+100x+5x^2+50x+125-64=0\)

\(4x^3+45x^2+150x+61=0\)

\(\Delta=b^2-3ac=45^2-3.4.150=225>1\) vậy phương trình có 1 nghiệm 

\(k=\dfrac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^3}}=\dfrac{637}{125}\)

\(x=\dfrac{\sqrt{\Delta}.\left|k\right|}{3ak}.\left(\sqrt[3]{\left|k\right|+\sqrt{k^2+1}}+\sqrt[3]{\left|k\right|-\sqrt{k^2-1}}\right)-\dfrac{b}{3a}=-0,4702284538\)

3.\(\left(x+10\right)^2:\left(-2\right)=\left(-32\right)\Rightarrow\left(x+10\right)^2=16\)

\(\Rightarrow\left\{{}\begin{matrix}x+10=-4\\x+10=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-14\\x=-6\end{matrix}\right.\)

4.\(\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right)=0\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)