\(\dfrac{x}{18}=\dfrac{4}{3}\Rightarrow x=\dfrac{18.4}{3}=24\\ \dfrac{20}{y}=\dfrac{4}{3}\Rightarrow y=\dfrac{20.3}{4}=15\\ \dfrac{z}{21}=\dfrac{4}{3}\Rightarrow z=\dfrac{21.4}{3}=28\)

Ta có:

\(\dfrac{x}{18}\) = \(\dfrac{4}{3}\)

⇒ x = \(\dfrac{4}{3}\) . 18

⇒ x = 24

\(\dfrac{20}{y}\) = \(\dfrac{4}{3}\)

⇒ y = 20 : \(\dfrac{4}{3}\)

⇒ y = 15

\(\dfrac{z}{21}\) = \(\dfrac{4}{3}\)

⇒ z = \(\dfrac{4}{3}\) . 21

⇒ z = 28

⇒ x + y + z = 24 + 15 + 28 = 67

Vậy x + y + z = 67

x = 24, y = 15, z = 28;

Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\)

nên \(\dfrac{x}{7}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{z}=\dfrac{5}{8}\)

nên \(\dfrac{y}{5}=\dfrac{z}{8}\)

hay \(\dfrac{y}{20}=\dfrac{z}{32}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

hay \(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

mà 2x-5y+2z=100

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

\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x-5y+2z}{14-100+64}=\dfrac{100}{-22}=\dfrac{-50}{11}\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-50}{11}\\\dfrac{y}{20}=\dfrac{-50}{11}\\\dfrac{z}{32}=-\dfrac{50}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{350}{11}\\y=\dfrac{-1000}{11}\\z=\dfrac{-1600}{11}\end{matrix}\right.\)

Ta có:  \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\Rightarrow\dfrac{x}{14}=\dfrac{y}{40}\Rightarrow\dfrac{2x}{28}=\dfrac{5y}{200}\) \(\left(1\right)\)

Lại có:  \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{40}=\dfrac{z}{64}\Rightarrow\dfrac{5y}{200}=\dfrac{2z}{128}\)   \(\left(2\right)\)

Kết hợp ( 1 ) và ( 2 ) ta có:     \(\dfrac{2x+5y-2z}{28+200-128}=\dfrac{100}{100}=1\)

⇒  \(\dfrac{2x}{28}=1\Rightarrow x=\dfrac{1.28}{2}=14\)

⇒  \(\dfrac{5y}{200}=1\Rightarrow y=\dfrac{1.200}{5}=40\)

⇒  \(\dfrac{2z}{128}=1\Rightarrow z=\dfrac{1.128}{2}=64\)

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x + ( x + 1 ) + ( x + 2 ) +...+ 19 + 20 = 20

= x + ( x + 1 ) + ( x + 2 ) + ...+ ( -2 ) + ( - 1 ) + 0 + 1 + 2 + ..

=> x + ( x + 1 ) + ( x + 2 ) + ...+ ( -2 ) + ( -1 ) + 0 + 1 +. .. + 19...

do đó vế trái là tổng các cặp số đối

vậy x = ( -19 )

k mk nha

mình làm xong bài đấy lâu rùi

a) \(4x-2=x\)

\(4x-x=2\)

\(3x=2\)

\(x=\dfrac{2}{3}\)

b) Thay \(x=1,y=3\) ta có \(3=a.1\Rightarrow a=3\)

Vậy hàm số cần tìm là \(y=3x\)

c) Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{x+y+z}{1+2+3}=\dfrac{180}{6}=30\)

\(\Rightarrow\left\{{}\begin{matrix}x=30\times1=30\\y=30\times2=60\\z=30\times3=90\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau, có:

    \(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{20}{5}=4\)

Suy ra: \(\frac{x}{2}=4\Rightarrow x=4\cdot2=8\)

              \(\frac{y}{3}=4\Rightarrow y=3\cdot4=12\)

=>20x+190=950

=>20x=760

hay x=38

`20x+190=950`

`20x=760`

`x= 760: 20`

`x= 38`

20 x X = 950 - (1 + 2 + 3 + 4 + 5 + ... + 19)

20 x X = 950 - 190

20 x X = 760

X = 760 : 20

X = 38

<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)

<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)

<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)

Vì \(x^2+x+5>0\)=>x-4=0

<=>x=4

Bài 4:

a: xy=-2

=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)

=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)

b: \(\left(x-1\right)\left(y+2\right)=-3\)

=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)

=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)

Bài 3:

a: \(x\left(x+9\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b: \(\left(x-5\right)^2=9\)

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

=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)

c: \(\left(7-x\right)^2=-64\)

mà \(\left(7-x\right)^2>=0\forall x\)

nên \(x\in\varnothing\)

Bài 2:

a: \(\left(-31\right)\cdot x=-93\)

=>\(31\cdot x=93\)

=>\(x=\dfrac{93}{31}=3\)

b: \(\left(-4\right)\cdot x=-20\)

=>\(4\cdot x=20\)

=>\(x=\dfrac{20}{4}=5\)

c: \(5x+1=-4\)

=>\(5x=-4-1=-5\)

=>\(x=-\dfrac{5}{5}=-1\)

d: \(-12x+1=-4\)

=>\(-12x=-4-1=-5\)

=>\(12x=5\)

=>\(x=\dfrac{5}{12}\)