a)

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\\ \Leftrightarrow\frac{6x+30}{24}-\frac{16x-24}{24}-\frac{18x-3}{24}-\frac{4x-2}{24}=0\\ \Leftrightarrow\frac{6x+30-16x+24-18x+3-4x+2}{24}=0\\ \Leftrightarrow\frac{59-32x}{24}=0\\ \Rightarrow59-32x=0\\ \Rightarrow x=\frac{59}{32}\)

b)

\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\\ \Leftrightarrow\frac{6x+24-30x+120-10x+15x-30}{30}=0\\ \Leftrightarrow\frac{114-19x}{30}=0\\ \Rightarrow114-19x=0\\ \Rightarrow x=\frac{-144}{-19}=6\\ \Rightarrow x=6\)

c)

\(x^2-3x+2=0\\ \Leftrightarrow2-x-2x+x^2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

a. \(1+\frac{2x-5}{6}=\frac{3-x}{4}\)

\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}-\frac{3\left(3-x\right)}{12}=0\)

\(\Leftrightarrow12+4x-10-9+3x=0\)

\(\Leftrightarrow7x-7=0\)

\(\Leftrightarrow7x=7\Leftrightarrow x=1\)

b. \(\frac{x+1}{2}-\frac{x-2}{3}=\frac{3\left(x+1\right)}{6}-\frac{2\left(x-2\right)}{6}=\frac{x+7}{6}\)

c. \(\frac{2x-1}{3}+x=\frac{x+4}{2}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}+\frac{6x}{6}-\frac{3\left(x+4\right)}{6}=0\)

\(\Leftrightarrow7x-14=0\)

\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

d. \(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}+\frac{2x-1}{12}\)

\(=\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}-\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)

\(=6x+30-16x+24-18x+3+4x-2\)

\(=-24x-55\)

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha , 

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

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

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)