\(x\) = 0 - \(x\)

\(x\) + \(x\) = 0

2\(x\)       = 0

   \(x\)       = 0

ko  biết

\(\text{70,2 : 9 = 7,8}\)

70,2:9=7,8

2^20 chứ không phải là 26^20 đâu nhé

2                                   3                                   4

3                                   4                                   2 

4                                   2                                   3

\(2x^2y+4xy^2+2y^3-8\) 

\(=2y\left(x^2+2xy+y^2\right)-8\)

\(=2y\left(x+y\right)^2-8\) 

\(=2\left[y\left(x+y\right)^2-4\right]\)

thank you huỳnh thanh phong

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

\(=\left(x^2+2\cdot5\cdot x+5^2\right)-\left(x^2+3x-2x-6\right)\)

\(=\left(x^2+10x+25\right)-\left(x^2+x-6\right)\)

\(=x^2+10x+25-x^2-x+6\)

\(=9x+31\)

\((x+5)^2-(x+3)(x-2)\\=(x^2+2\cdot x\cdot5+5^2)-[x(x-2)+3(x-2)]\\=(x^2+10x+25)-(x^2-2x+3x-6)\\=x^2+10x+25-(x^2+x-6)\\=x^2+10x+25-x^2-x+6\\=(x^2-x^2)+(10x-x)+(25+6)\\=9x+31\)

\(-4x^2-24xy-36y^2\)

\(=-\left(4x^2+24xy+36y^2\right)\)

\(=-\left[\left(2x\right)^2+24xy+\left(6y\right)^2\right]\)

\(=-\left[\left(2x\right)^2+2\cdot2x\cdot6y+\left(6y\right)^2\right]\)

\(=-\left(2x+6y\right)^2\)

\(=-\left[2\left(x+3y\right)\right]^2\)

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

Chịu nhá

\(3x=2y=5z\)

\(\Rightarrow\dfrac{3x}{30}=\dfrac{2y}{30}=\dfrac{5z}{30}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+y+z}{10+15+6}=\dfrac{-62}{31}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=-2\\\dfrac{y}{15}=-2\\\dfrac{z}{6}=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\cdot10=-20\\y=-2\cdot15=-30\\y=-2\cdot6=-12\end{matrix}\right.\)

Ta có: \(3x=2y=5z\)

\(\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{2}}=\dfrac{z}{\dfrac{1}{5}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau và \(x+y+z=-62\), ta được:

\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{2}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x+y+z}{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{5}}=\dfrac{-62}{\dfrac{31}{30}}=-60\)

\(\Rightarrow\left\{{}\begin{matrix}x=-60\cdot\dfrac{1}{3}=-20\\y=-60\cdot\dfrac{1}{2}=-30\\z=-60\cdot\dfrac{1}{5}=-12\end{matrix}\right.\)

Vậy \(x=-20;y=-30;z=-12\).

\(5^3-3\left(x-3\right)=25\)

\(\Rightarrow125-3\left(x-3\right)=25\)

\(\Rightarrow3\left(x-3\right)=125-25\)

\(\Rightarrow3\left(x-3\right)=100\)

\(\Rightarrow x-3=\dfrac{100}{3}\)

\(\Rightarrow x=\dfrac{100}{3}+3=\dfrac{100}{3}+\dfrac{9}{3}\)

\(\Rightarrow x=\dfrac{109}{3}\)

\(#WendyDang\)

$5^3-3\left(x-3\right)=25$

⇒125−3(�−3)=25$\Rightarrow 125-3\left(x-3\right)=25$

⇒3(�−3)=125−25$\Rightarrow 3\left(x-3\right)=125-25$

⇒3(�−3)=100$\Rightarrow 3\left(x-3\right)=100$

⇒�−3=1003$\Rightarrow x-3=\genfrac{}{}{0ex}{}{100}{3}$

⇒�=1003+3=1003+93$\Rightarrow x=\genfrac{}{}{0ex}{}{100}{3}+3=\genfrac{}{}{0ex}{}{100}{3}+\genfrac{}{}{0ex}{}{9}{3}$

⇒�=1093$\Rightarrow x=\genfrac{}{}{0ex}{}{109}{3}$

Đúng cholike

Tui k bt gì né

1-2+3-4+5-6+...+99-100+101

= (1+3+5+...+101) - (2+4+6+...+100)

từ 1 đến 101 có : (101-1):2+1=51

1+..+101 = (1+101)x 51:2= 2601

Từ 2 đến 100 có : (100-2);2+1=50

2+...+100 = (100 +2) x 50:2=2550

=> 2601-2550=51

#Họctốt

NHWOS TICK NHA :D