\(\Leftrightarrow\frac{2^{3x^2-3x+1}}{3^{x^2-x+1}}.\frac{3^{2x^2-3x+2}}{5^{2x^2-3x+2}}.\frac{5^{3x^2-4x+3}}{7^{3x^2-4x+3}}.\frac{7^{4x^2-5x+4}}{2^{4x^2-5x+4}}=210^{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{\left(3.5.7\right)^{x^2-x+1}}{2^{x^2-2x+1}}=2^{\left(x-1\right)^2}.\left(3.5.7\right)^{\left(x-1\right)^2}\)

\(\Leftrightarrow105^x=2^{2\left(x-1\right)^2}\)

Lấy Logarit cơ số 2 hai vế, ta được :

\(2\left(x-1\right)^2=\left(\log_2105\right)x\)

\(\Leftrightarrow2x^2-\left(4+\log_2105\right)x+2=0\)

\(\Leftrightarrow x=\frac{\left(2+\log_2105\right)\pm\sqrt{\log^2_2105+8\log_2105}}{4}\)

Vậy phương trình đã cho có 2 nghiệm

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

Bài 1: 

b) Phương trình đã cho tương đương với phương trình:

\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)

\(\Leftrightarrow44x=-1056\)

\(\Leftrightarrow x=-24\)

Vậy x=-24 là nghiệm của phương trình

c) Phương trình đã cho tương đương với phương trình:

\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)

\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)

\(\Leftrightarrow-447x=894\)

\(\Leftrightarrow x=-2\)

Vậy x=-2 là nghiệm của phương trình

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)

\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)

\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)

\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)

a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)

b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)

c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)

a) 0,75x(x + 5) = (x + 5)(3 - 1,25x)

<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = (x + 5)(3 - 1,25x) - (x + 5)(3 - 1,25x)

<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = 0

<=> (x + 5)(0,75 + 1,25x - 3) = 0

<=> (x + 5)(2x - 3) = 0

<=> x + 5 = 0 hoặc 2x - 3 = 0

<=> x = -5 hoặc x = 3/2

b) 4/5 - 3 = 1/5x(4x - 15)

<=> -11/5 = x(4x - 15)/5

<=> -11 = x(4x - 15)

<=> -11 = 4x² - 15x

<=> 11 + 4x² - 15x = 0 

<=> 4x² - 4x - 11x + 11 = 0

<=> 4x(x - 1) - 11(x - 1) = 0

<=> (4x - 11)(x - 1) = 0

<=> 4x - 11 = 0 hoặc x - 1 = 0

<=> x = 11/4 hoặc x = 1

c) \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

<=> 12x - 36 - 2(x - 3)(2x - 5) = 3(x - 3)(3 - x)

<=> 12x - 36 - 4x² + 10x + 12x - 30 = 9x - 3x² - 27 + 9x

<=> 34x - 66 - 4x² = 18x - 3x² - 27

<=> 34x - 66 - 4x² - 18x + 3x² + 27 = 0

<=> 16x - 39x - x² = 0

<=> x² - 16x + 39x = 0

<=> (x - 3)(x - 13) = 0

<=> x - 3 = 0 hoặc x - 13 = 0

<=> x = 3 hoặc x = 13

d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

<=> (3x + 1)(3x - 2) + 15(3x + 1) = 2(2x + 1)(3x + 1) + 6x(3x + 1)

<=> 9x² - 6x + 3x - 2 + 45x + 15 = 12x³ + 4x + 6x + 2 + 18x² + 6x

<=> 9x² + 42x + 13 = 30x² + 16x + 2

<=> 9x² + 42x + 13 - 30x² - 16x - 2 = 0

<=> -21x² + 26x + 11 = 0

<=> 21x² - 26x - 11 = 0

<=> 21x² + 7x - 33x - 11 = 0

<=> 7x(3x + 1) - 11(3x + 1) = 0

<=> (7x - 11)(3x + 1) = 0

<=> 7x - 11 = 0 hoặc 3x + 1 = 0

<=> x = 11/7 hoặc x = -1/3