\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...

câu c sao tính ra vậy đc vậy k hiểu giải thích hộ e đi 36 đâu mất òi

b. \(\left(2x+1\right)+\left(4x+3\right)+\left(6x+5\right)+...+\left(100x+99\right)=7600\)

\(\rightarrow\left(2x+4x+6x+...+100x\right)+\left(1+3+5+...+99\right)=7600\)

\(\rightarrow\frac{\left(2x+100x\right).50}{2}+\frac{\left(1+99\right).50}{2}=7600\)

\(\rightarrow51x.50+50.50=7600\)

\(\rightarrow51x.50+2500=7600\)

\(\rightarrow51x.50=7600-2500\)

\(\rightarrow51x.50=5100\)

\(\rightarrow50x=100\)

\(\rightarrow x=\frac{100}{50}=2\)

Vậy x = 2

a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)

\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)

\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)

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

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

\(6x\left(-3x+4\right)=0\)

\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)

*) \(6x=0\)

\(x=0\)

*) \(-3x+4=0\)

\(3x=4\)

\(x=\dfrac{4}{3}\)

Vậy \(x=0;x=\dfrac{4}{3}\)

b) \(4x\left(x-2019\right)-x+2019=0\)

\(4x\left(x-2019\right)-\left(x-2019\right)=0\)

\(\left(x-2019\right)\left(4x-1\right)=0\)

\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)

*) \(x-2019=0\)

\(x=2019\)

*) \(4x-1=0\)

\(4x=1\)

\(x=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4};x=2019\)

1, ( x - 3 )¹⁰⁰ - 1 = 0 => ( x - 3 )¹⁰⁰ = 0 + 1 = 1

Mà 1 = 1¹⁰⁰ => x - 3 = 1 => x = 1 + 3 = 4 hoặc 1 = (-1)¹⁰⁰ => x - 3 = -1 => x = -1 + 3 = 2

2, ( 9 - 5x )²⁰¹⁹ + 1 = 0 => ( 9 - 5x )²⁰¹⁹ = 0 - 1 = -1

Mà -1 = (-1)²⁰¹⁹ => 9 - 5x = -1 => 5x = 9 - ( -1 ) = 10 => x = 10 : 5 = 2

3, ( 4x - 3 )⁵ = ( 2 - x )⁵ => 4x - 3 = 2 - x

=> 4x + x = 3 + 2 => 5x = 5 => x = 5 : 5 = 1

4, ( 2x - 9 )² = ( 5x - 6 )² => 2x - 9 = 5x - 6 ... ( tự làm )

5, ( 11 - 4x )⁶ - ( 2 - 5x )⁶ = 0 => ( 11- 4x )⁶ = ( 2 - 5x )⁶

=> 11 - 4x = 2 - 5x _ Đến đây làm tương tự 2 câu trên

6, ( x - 9 )⁹ = ( x - 9 )⁷ mà cơ số bằng nhau ( = x - 9 )

=> x - 9 = 1 hoặc -1 vì 1⁹ = 1⁷ và ( -1 )⁹ = ( -1 )⁷

TH1: x - 9 = 1 => x = 1 + 9 = 10

TH2: x - 9 = -1 => x = -1 + 9 = 8

7, 8, 9 tương tự 6 ( kết quả của cơ số đều = 1 hoặc -1 )

`@` `\text {Ans}`

`\downarrow`

`c)`

`(34 - 2x)(2x - 6) = 0`

`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

Vậy, `x \in {17; 3}`

`d)`

`(2019 - x)(3x - 12) = 0`

`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

Vậy, `x \in {2019; 4}.`

`@` `\text {Kaizuu lv uuu}`

a, x²(x - 3) + 12 - 4x = 0

<=> x²(x - 3) + 4(3 - x) = 0

<=> x²(x - 3) - 4(x - 3) = 0

<=> (x - 3)(x² - 4) = 0

<=> x - 3 = 0    hoặc   x² - 4 = 0

<=> x = 3                    x² = 4

<=> x = 3                    x = 2 hoặc x = -2

b, 2(x + 5) - x² - 5x = 0

<=>  2(x + 5) - x(x + 5) = 0

<=> (x + 5)(2 - x) = 0

<=> x + 5 = 0   hoặc 2 - x = 0

<=> x = -5                  x = 2

c, 2x(x + 2019) - x - 2019 = 0

<=> 2x(x + 2019) - (x + 2019) = 0

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

<=> x + 2019 = 0  hoặc  2x - 1 = 0

<=> x = -2019                 2x = 1

<=> x = -2019                  x = 1/2

a,\(\text{Để }\left(4x+3\right)^3-\left(2x-5\right)^3=\left(2x+8\right)^3\) thì

\(3\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\2x-5=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy..

b,\(Để\left(3x+2016\right)^3+\left(3x-2019\right)^3=\left(6x-3\right)^3\) thì

\(3\left(3x+2016\right)\left(3x-2019\right)\left(6x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2016=0\\3x-2019=0\\6x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2016}{3}\\x=\dfrac{2019}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy...