\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{50}\end{matrix}\right.\)

\(\Leftrightarrow\left(x-y\right)^2=\dfrac{3}{10}+\dfrac{3}{50}=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y=\dfrac{3}{5}\\x-y=-\dfrac{3}{5}\end{matrix}\right.\)

thay vào tìm x và y

Ta có: x(x-y)=3/10.

y(x-y)=-3/50.

=>x(x-y)-y(x-y)=3/10-(-3/50).

=>(x-y)^2=9/25=(±3/5)^2.

=>x-y=±3/5.

+)x-y=3/5.

=>x=3/10:3/5=1/2.

=>y=1/2-3/5=-1/10.

+)x-y=-3/5.

=>x=3/10:(-3/5)=-1/2.

=>y=-1/2-(-3/5)=1/10.

Vậy:x=±1/2;y=±1/10.

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

a: \(\left|x-3\right|\ge0\forall x\)

\(\left|2y-6\right|\ge0\forall y\)

Do đó: \(\left|x-3\right|+\left|2y-6\right|\ge0\forall x,y\)

=>\(\left|x-3\right|+\left|2y-6\right|+10\ge10\forall x,y\)

\(\left(y-3\right)^2\ge0\forall y\)

=>\(\left(y-3\right)^2+3\ge3\forall y\)

=>\(\frac{30}{\left(y-3\right)^2+3}\le\frac{30}{3}=10\forall y\)

Ta có: \(\left|x-3\right|+\left|2y-6\right|+10=\frac{30}{\left(y-3\right)^2+3}\)

mà \(\left|x-3\right|+\left|2y-6\right|+10\ge10\forall x,y\)

và \(\frac{30}{\left(y-3\right)^2+3}\le\frac{30}{3}=10\forall y\)

nên dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3

b: \(\left(2x+6\right)^{2020}\ge0\forall x\)

=>\(\left(2x+6\right)^{2020}+51\ge51\forall x\)

Ta có: \(\left|x+3\right|\ge0\forall x\)

=>\(3\left|x+3\right|\ge0\forall x\)

=>\(3\left|x+3\right|+2\ge2\forall x\)

=>\(\frac{102}{3\left|x+3\right|+2}\le\frac{102}{2}=51\forall x\)

Ta có: \(\left(2x+6\right)^{2020}+51=\frac{102}{3\left|x+3\right|+2}\)

mà \(\left(2x+6\right)^{2020}+51\ge51\forall x\)

và \(\frac{102}{3\left|x+3\right|+2}\le\frac{102}{2}=51\forall x\)

nên dấu '=' xảy ra khi x+3=0

=>x=-3

Bài 1:

\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)

Bài  2:

\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)

a: ĐKXĐ: \(x^2+y^2\ne0\)

=>\(\left[{}\begin{matrix}x^2\ne0\\y^2\ne0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

b: ĐKXĐ: \(x^2-2x+1\ne0\)

=>\(\left(x-1\right)^2\ne0\)

=>\(x-1\ne0\)

=>\(x\ne1\)

c: ĐKXĐ: \(x^2+6x+10\ne0\)

=>\(x^2+6x+9+1\ne0\)

=>\(\left(x+3\right)^2+1\ne0\)(luôn đúng)

d:ĐKXĐ: \(\left(x+3\right)^2+\left(y-2\right)^2\ne0\)

=>\(\left[{}\begin{matrix}x+3\ne0\\y-2\ne0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x\ne-3\\y\ne2\end{matrix}\right.\)

1: x:y:z=3:5:(-2)

=>\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)

mà 5x-y+3z=-16

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=\frac{5x-y+3z}{5\cdot3-5+3\cdot\left(-2\right)}=\frac{-16}{15-5-6}=\frac{-16}{10-6}=\frac{-16}{4}=-4\)

=>\(\begin{cases}x=-4\cdot3=-12\\ y=-4\cdot5=-20\\ z=\left(-4\right)\cdot\left(-2\right)=8\end{cases}\)

2: \(\frac{x}{2}=\frac{y}{-3}\)

=>\(\frac{x}{-2}=\frac{y}{3}\)

=>\(\frac{x}{-8}=\frac{y}{12}\) (1)

\(\frac{y}{4}=\frac{z}{3}\)

=>\(\frac{y}{12}=\frac{z}{9}\) (2)

Từ (1),(2) suy ra \(\frac{x}{-8}=\frac{y}{12}=\frac{z}{9}\)

mà x+y+z=5,2

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{-8}=\frac{y}{12}=\frac{z}{9}=\frac{x+y+z}{-8+12+9}=\frac{5.2}{13}=0,4\)

=>\(\begin{cases}x=-8\cdot0,4=-3,2\\ y=12\cdot0,4=4,8\\ z=9\cdot0,4=3,6\end{cases}\)

3: 2x=3y

=>\(\frac{x}{3}=\frac{y}{2}\)

=>\(\frac{x}{21}=\frac{y}{14}\)

7z=5y

=>\(\frac{z}{5}=\frac{y}{7}\)

=>\(\frac{y}{14}=\frac{z}{10}\)

=>\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

mà 3x-7y+5z=30

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{3\cdot21-7\cdot14+5\cdot10}=\frac{30}{63-98+50}=\frac{30}{63-48}=\frac{30}{15}=2\)

=>\(\begin{cases}x=2\cdot21=42\\ y=2\cdot14=28\\ z=2\cdot10=20\end{cases}\)

4: 3x=4y=5z

=>\(\frac{3x}{60}=\frac{4y}{60}=\frac{5z}{60}\)

=>\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}\)

mà x-(y+z)=-21

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}=\frac{x-\left(y+z\right)}{20-\left(15+12\right)}=\frac{-21}{20-\left(27\right)}=\frac{-21}{-7}=3\)

=>\(\begin{cases}x=3\cdot20=60\\ y=3\cdot15=45\\ z=3\cdot12=36\end{cases}\)

5: Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

=>\(\begin{cases}x-1=2k\\ y-2=3k\\ z-3=4k\end{cases}\Rightarrow\begin{cases}x=2k+1\\ y=3k+2\\ z=4k+3\end{cases}\)

2x+3y-z=50

=>2(2k+1)+3(3k+2)-(4k+3)=50

=>4k+2+9k+6-4k-3=50

=>9k+5=50

=>9k=45

=>k=5

=>\(\begin{cases}x=2\cdot5+1=11\\ y=3\cdot5+2=15+2=17\\ z=4\cdot5+3=20+3=23\end{cases}\)

Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)

Tương tự:

\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)

\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)

\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)

Ta có: \(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)

\(\left|y-2\right|+\left|y+2\right|=\left|2-y\right|+\left|y+2\right|\ge\left|2-y+y+2\right|=4\)

\(\Rightarrow\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{16}{4}=4\Rightarrow\left|x+3\right|+\left|x-1\right|\ge\dfrac{6}{\left|y-2\right|+\left|y+2\right|}\)

Dấu '=' xảy ra <=> (x+3)(1-x)\(\ge0\) và (2-y)(y+2)\(\ge0\)

Vì x,y \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)