ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

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

Vậy: S={-4;-5}

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

GHI RÕ DÙM MÌNH ĐK CỦA CẢ 3 CÂU LUÔN ĐC KO Á.

1: 2*2*2*2*2*2=2^6

2: x*x*x*x=x^4

3: =7^4*6^4=42^4

4: =5^2*3^2*4^2=60^2

5: =2*3^2*5^5

6: =10^3*10^3=10^6

Không biết nãy bị lỗi ở đâu, mình gửi lại:<

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

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

vậy x=0 và x=2

3x(x-2)-x^2+2x=0

<=>3x(x-2)-x(x-2)=0

<=>(3x-x)(x-2)=0

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

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2

Hướng làm:

Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức

\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)

\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)

\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)

\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)

\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)

\(< =>x+2021=0< =>x=-2021\)

Vậy....

x+1000=7896

      x    =7896-1000

     x     =6896

tk mk nha mn

x=6896

1000 : [30 + (2^x - 6)] = 3² + 4²

1000 : [30 + (2^x - 6)] = 9 + 16

1000 : [30 + (2^x - 6)] = 25

30 + (2^x - 6) = 1000 : 25

30 + (2^x - 6) = 40

2^x - 6 = 40 - 30

2^x - 6 = 10

2^x = 10 + 6

2^x = 16

2^x = 2⁴

⇒ x = 4

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

sao câu 1 hoài v ạ.Còn câu 2,3 nữa á.