Thanks

Bài 1

a) (x + 3)(x + 2) = 0

x + 3 = 0 hoặc x + 2 = 0

*) x + 3 = 0

x = 0 - 3

x = -3 (nhận)

*) x + 2 = 0

x = 0 - 2

x = -2 (nhận)

Vậy x = -3; x = -2

b) (7 - x)³ = -8

(7 - x)³ = (-2)³

7 - x = -2

x = 7 + 2

x = 9 (nhận)

Vậy x = 9

Bài 3

20a + 10b = 2010

10b = 2010 - 20a

b = (2010 - 20a) : 10

*) a = 0

b = (2010 - 20.0) : 10 = 201

*) a = 1

b = (2010 - 10.1) : 10 = 200

*) a = 2

b = (2010 - 10.2) : 10 = 199

Vậy ta có ba cặp số nguyên (a; b) thỏa mãn:

(0; 201); (1; 200); (2; 199)

a)\(\left|2x-3y\right|+\left|2y-4z\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\forall x;y\\\left|2y-4z\right|\ge0\forall y;z\end{matrix}\right.\) \(\Rightarrow\left|2x-3y\right|+\left|2y-4z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|2y-4z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2y=4z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{y}{4}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}\)

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

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x+y+z}{6+4+2}=\dfrac{7}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{12}.6=\dfrac{7}{2}\\y=\dfrac{7}{12}.4=\dfrac{7}{3}\\z=\dfrac{7}{12}.2=\dfrac{7}{6}\end{matrix}\right.\)

b)\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x-3\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x-3\right|=0\\\left|x-4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)

Vì \(2\ne3\ne4\) nên \(x\in\varnothing\)

c)

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+8\right|+\left|x+9\right|\)

Với mọi \(x\ge0\) ta có:

\(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+8\right|=x+8\\\left|x+9\right|=x+9\end{matrix}\right.\)\(\Leftrightarrow x+1+x+2+...+x+8+x+9=x-1\)

\(\Leftrightarrow9x+90=x-1\)

\(\Leftrightarrow9x=x-89\)

\(\Leftrightarrow-8x=89\)

\(\Leftrightarrow x=\dfrac{89}{-8}\left(KTM\right)\)

Với mọi \(x< 0\) ta có:

\(\left\{{}\begin{matrix}x+1=-x-1\\x+2=-x-2\\x+8=-x-8\\x+9=-x-9\end{matrix}\right.\) \(\Leftrightarrow\left(-x-1\right)+\left(-x-2\right)+...+\left(-x-8\right)+\left(-x-9\right)=x-1\)

\(\Leftrightarrow-9x-90=x-1\)

\(\Leftrightarrow-9x=x+89\)

\(\Leftrightarrow-10x=89\)

\(\Leftrightarrow x=\dfrac{89}{-10}\left(TM\right)\)

d)\(\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\\ \left|5y-2z\right|\ge0\\ \left|2z-6\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|5y-2z\right|=0\\\left|2z-6\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=3\\y=\dfrac{6}{5}\\x=\dfrac{9}{5}\end{matrix}\right.\)

\(\frac{12}{-6}=-2=\frac{-10}{5}=\frac{-6}{3}=\frac{34}{-17}=\frac{-18}{9}\)

Vậy...........

\(\frac{-24}{-6}=4=\frac{12}{3}=\frac{4}{\left(\pm1\right)^2}=\frac{\left(-2\right)^3}{-2}\)

Vậy......

Sử dụng AM - GM dạng cộng mẫu :

\(\frac{1}{x+1}+\frac{4}{y+2}+\frac{9}{z+3}\)

\(\ge\frac{\left(1+2+3\right)^2}{x+y+z+1+2+3}\)

\(=\frac{36}{x+y+z+6}\)

\(=\frac{36}{12}=3\)

Đẳng thức xảy ra tại ......

Trên kia là sai lầm thường gawpjjj ( theo mình nghĩ thế tại nhác tìm dấu bằng )

thứ 2 là wolfram alpha bảo không có minimize:

a, \(x:y:z=2:3:4\&x+y+z=365\)

\(x:y:z=2:3:4\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tích chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{365}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{365}{9}\\\dfrac{y}{3}=\dfrac{365}{9}\\\dfrac{z}{4}=\dfrac{365}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{730}{9}\\y=\dfrac{365}{3}\\z=\dfrac{1460}{9}\end{matrix}\right.\)

b:\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\\dfrac{7}{2}+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)

c: =>1/2x-5=0 và y^2-1/4=0

=>\(\left\{{}\begin{matrix}x=10\\y\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\end{matrix}\right.\)

d: =>x=0 và y-1/10=0

=>x=0 và y=1/10