\(x+\frac{2}{15}=\frac{1}{3}\)

\(x=\frac{1}{3}-\frac{2}{15}\)

\(x=\frac{1}{5}\)

h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)

\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)

\(\frac{2}{3}:x=\frac{1}{12}\)

\(x=\frac{2}{3}:\frac{1}{12}\)

\(x=8\)

x3 , y3 là gì vậy bn ???

X= \(\frac{45}{7}\)

`1)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)`

`<=>2x^2-5x-12+x^2-7x+10=3x^2-17x+20`

`<=>3x^2-12x-2=3x^2-17x+20`

`<=>5x=22`

`<=>x=22/5`

Vậy `S={22/5}`

`2)x^2(x-2019)=2019-x`

`<=>(x-2019)(x^2+1)=0`

`<=>x-2019=0`

`<=>x=2019(do \ x^2+1>=1>0)`

Vậy `S={2019}`

\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|+\left|x-5\right|\)

\(\Rightarrow A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|4-x\right|+\left|5-x\right|\)

\(\Rightarrow A\ge x-1+x-2+0+4-x+5-x\)

\(\Rightarrow A\ge6\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-1\ge0;x-2\ge0\\x-3=0\\x-4\le0;x-5\le0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ge2\\x=3\\x\le4\end{cases}}\Rightarrow x\in\left(2;3;4\right)\)

Vậy MinA = 6 \(\Leftrightarrow x\in\left(2,3,4\right)\)

Phạm Tuấn Đạt dùng lý thuyết nào vậy?

Hơi tắt quá, chi tiết 1 chút đc ko ?

(2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
<=>4x^2-4x+1+x^2+6x+9-5x^2+245=0
<=>2x+255=0
<=>2x=-255
<=>x=-255/2

Có trên google ( ghi nguồn đầy đủ )

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\)\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow\)\(5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow\)\(2x=-255\)

\(\Leftrightarrow\)\(x=-127,5\)

Vậy...

4x^2 - 2.2x.1 + 1^2 + x^2 + 2.x.3 + 3^2 - 5.x^2 + 5. 7^2 = 0

4x^2 - 4x + 1 + x^2 + 6x + 9 - 5x^2 + 245 = 0

2x + 1 + 9 + 245 = 0

=> x = 127,5

a: \(A=\dfrac{x+2}{x}\cdot\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

b: \(B=\dfrac{2x-6+3x+9-5x+2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5}{x^2-9}\)