`Answer:`

`\frac{x-1}{34}+\frac{x-2}{34}=\frac{x-3}{32}+\frac{x-4}{31}`

`<=>\frac{x-1}{34}-1+\frac{x-2}{33}-1=\frac{x-3}{32}-1+\frac{x-4}{31}-1`

`<=>\frac{x-35}{34}+\frac{x-35}{33}=\frac{x-35}{32}+\frac{x-35}{31}`

`<=>(x-35)(\frac{1}{34}+\frac{1}{33}-\frac{1}{32}-\frac{1}{31})=0`

`<=>x-35=0`

`<=>x=35`

a)

 \(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\left(x-2\right)\\ \Leftrightarrow\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)-\left(1-x\right)\left(x-2\right)=0\\ 6x-2x^2+x-3+x^2+3x-2x-6-x+2+x^2-2x=0\) 

   \(\Leftrightarrow5x-7=0\\ \Leftrightarrow5x=7\Leftrightarrow x=\frac{7}{5}\)

Vây x=7/5

câu b tương tự vậy nhưng mak mình lười lm quá

3x⁴ + x² - 4 = 0

Đặt t = x² ( a ≥ 0 ) pt đã cho trở thành 3t² + t - 4 = 0 

Dễ thấy pt trên có a + b + c = 0 nên có hai nghiệm t₁ = 1 (nhận) ; t₂ = c/a = -4/3 (loại)

=> x² = 1 <=> x = ±1

1. \(x-8+32=68\)

\(x-8=68-32\)

\(x-8=36\)

\(x=36+8\)

\(x=44\)

_____

2. \(x+8+32=68\)

\(x+8=68-32\)

\(x+8=36\)

\(x=36-8\)

\(x=28\)

_____

3. \(98-x+34=43\)

\(98-x=43-34\)

\(98-x=9\)

\(x=98-9\)

\(x=89\)

_____

4. \(98+x-34=43\)

\(98+x=43+34\)

\(98+x=77\)

\(x=77-98\)

\(x=-21\)

_____

5. \(x:5:4=800\)

\(x:5=800.4\)

\(x:5=3200\)

\(x=3200.5\)

\(x=16000\)

_____

6. \(18+x=384:8\)

\(18+x=48\)

\(x=48-18\)

\(x=30\)

_____

7. \(\left(84,6-2\times x\right):3,02=5,1\)

\(84,6-2.x=5,1.3,02\)

\(84,6-2.x=15,402\)

\(2.x=84,6-15,402\)

\(2.x=69,198\)

\(x=69,198:2\)

\(x=34,599\)

_____

8. \(x\times5=120:6\)

\(x.5=20\)

\(x=20:5\)

\(x=4\)

_____

9. \(\left(15\times24-x\right):0,25=100:0,25\)

\(\left(15.24-x\right):0,25=400\)

\(15.24-x=400.0,25\)

\(15.24-x=100\)

\(360-x=100\)

\(x=360-100\)

\(x=260\)

1,x-8+32=68

=>x-8=68-32=36

=>x=36+8=42

2,x+8+32=68

=>x+8=68-32=36

=>x=36-8=28

3,98-x+34=43

=>98-x=43-34=9

=>x=98-9=89

4,98+x-34=43

=>98+x=43+34=77

=>x=77-98=-21

5,x:5:4=800

=>x:20=800

=>x=800x20=16000

6,18+x=384:8=48

=>x=48-18=30

7,(84,6-2x):3,02=5,1

=>84,6-2x=5,1x3,02=15,402

=>2x=84,6-15,402=69,198

=>x=69,198:2=34,599

8,5x=120:6=20

=>x=20:5=4

9,(15x24-x):0,25=100:0,25=400

=>360-x=400x0,25=100

=>x=360-100=260.

1. \(...\Rightarrow x+24=68\Rightarrow x=68-24=44\)

2. \(...\Rightarrow x+40=68\Rightarrow x=68-40=28\)

3. \(...\Rightarrow3^{2\left(8+x\right)}=9=3^2\Rightarrow2\left(8+x\right)=2\Rightarrow8+x=1\Rightarrow x=-7\)

4. \(...\Rightarrow3^{2\left(8+x\right)}=77\) (bạn xem lại đề)

5. \(...\Rightarrow x:5=800.4\Rightarrow x:5=3200\Rightarrow x=3200.5=16000\)

6. \(...\Rightarrow18+x=48\Rightarrow x=48-18=30\)

7. \(...\Rightarrow84,6-2x=5,1.3,02=15,402\Rightarrow2x=84,6-15,402\Rightarrow2x=69,198\Rightarrow x=34,599\)

8. \(...\Rightarrow5.x=20\Rightarrow x=20:5=4\)

9. \(\left(360-x\right):0,25=400\Rightarrow360-x=400.0,25=100\Rightarrow x=400-100=300\)

1: Tìm x

a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)

\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)

\(\Leftrightarrow3^x=3^{12}\cdot3^3\)

\(\Leftrightarrow3^x=3^{15}\)

hay x=15

Vậy: x=15

b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)

\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: x=2

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)

ai đó giúp mik vs :((

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(2x-1\right)^3=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=3^3\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)