Bài 1:

a) 5(x-3)-4=2(x-1)

\(\Leftrightarrow5x-15-4=2x-2\)

\(\Leftrightarrow5x-19-2x+2=0\)

\(\Leftrightarrow3x-17=0\)

\(\Leftrightarrow3x=17\)

\(\Leftrightarrow x=\frac{17}{3}\)

Vậy: \(x=\frac{17}{3}\)

b) 5-(6-x)=4(3-2x)

\(\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow-1+x-12+8x=0\)

\(\Leftrightarrow-13+9x=0\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\frac{13}{9}\)

Vậy: \(x=\frac{13}{9}\)

c) (3x+5)(2x+1)=(6x-2)(x-3)

\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow33x=1\)

\(\Leftrightarrow x=\frac{1}{33}\)

Vậy: \(x=\frac{1}{33}\)

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

\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)

\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)

\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow x=1\)

Vậy:x=1

Bài 2:

a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)

\(\Leftrightarrow4x-10x-15x-3x+60=0\)

\(\Leftrightarrow-24x+60=0\)

\(\Leftrightarrow-24x=-60\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy: \(x=\frac{5}{2}\)

b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)

\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)

\(\Leftrightarrow-3x=0\)

\(\Leftrightarrow x=0\)

Vậy: x=0

c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)

\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)

\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow3x+48=0\)

\(\Leftrightarrow3x=-48\)

\(\Leftrightarrow x=-16\)

Vậy: x=-16

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

\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)

\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)

\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)

\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)

\(\Leftrightarrow-13x+143=0\)

\(\Leftrightarrow-13x=-143\)

\(\Leftrightarrow x=11\)

Vậy: x=11

e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)

\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)

\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)

\(\Leftrightarrow45x-18-24-28x+60x-420=0\)

\(\Leftrightarrow77x-462=0\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

Vậy:x=6

Bài 3:

a) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)

Vì \(2\ne0\)

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

Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)

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

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

Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)

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

Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(2x+1=0\)

\(\Leftrightarrow2x=-1\)

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

Vậy: \(x=\frac{-1}{2}\)

d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)

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

Ta lại có \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)

Ta có: \(4\ne0\)(4)

Từ (3) và (4) suy ra

2x-1=0

\(\Leftrightarrow2x=1\)

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

Vậy: \(x=\frac{1}{2}\)

Bài 4:

a) \(\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)\)

\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)

\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)

\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)

\(\Leftrightarrow x^2+2x-8=0\)

\(\Leftrightarrow x^2+2x+1-9=0\)

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

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

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

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

Vậy: \(x\in\left\{2;-4\right\}\)

b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

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

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

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

\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)

Vì \(3\ne0\)

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

Vậy: \(x\in\left\{0;4\right\}\)

c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

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

\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)

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

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

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

Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)

d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)

\(\Leftrightarrow-8x^2+40x-32=0\)

\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)

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

Vì \(-8\ne0\)

nên \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{1;4\right\}\)

e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+23x+35x+115=0\)

\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)

Bài 5:

a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)

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

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)

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

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

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

Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)

b) \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)

\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)

\(\Leftrightarrow3x^2-3=0\)

\(\Leftrightarrow3\left(x^2-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)

Vì \(3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-1\right\}\)

c) \(x^4+x^3+x+1=0\)

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

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

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

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)

Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)

Từ (5) và (6) suy ra

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy: x=-1

ko khó đâu, chủ yếu nhát làm

Câu 1:

a.5.(x-3)-4=2.(x-1)

⇔5x-15-4=2x-2

⇔ 5x-2x=-2+19

⇔ 3x=17

⇔ x=17/3

b. 5-(6-x)=4.(3-2x)

⇔ x-1=12-8x

⇔ x+8x=12+1

⇔ x=13/9

c.(3x+5).(2x+1)=(6x-2).(x-3)

⇔ 6x² + 3x+10x+5=6x²-18x-2x+6

⇔ (6x²-6x²)+(13x+20x)=6-5

⇔ 33x=1

⇔x=1/33

d.(x+2)²+2.(x-4)=(x-4).(x-2)

⇔x²+4x+4+2x-8=x²-2x-4x+8

⇔(x²-x²)+(6x+6x)=8+8-4

⇔12x=12

⇔ x=1

a) ( x - 3)⁴ + ( x - 5)⁴ = 82

Đặt : x - 4 = a , ta có :

( a + 1)⁴ + ( a - 1)⁴ = 82

⇔ a⁴ + 4a³ + 6a² + 4a + 1 + a⁴ - 4a³ + 6a² - 4a + 1 = 82

⇔ 2a⁴ + 12a² - 80 = 0

⇔ 2( a⁴ + 6a² - 40) = 0

⇔ a⁴ - 4a² + 10a² - 40 = 0

⇔ a²( a² - 4) + 10( a² - 4) = 0

⇔ ( a² - 4)( a² + 10) = 0

Do : a² + 10 > 0

⇒ a² - 4 = 0

⇔ a = + - 2

+) Với : a = 2 , ta có :

x - 4 = 2

⇔ x = 6

+) Với : a = -2 , ta có :

x - 4 = -2

⇔ x = 2

KL.....

b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8

⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680

⇔ ( n² - 9n + 18)( n² - 9n + 20) = 1680

Đặt : n² - 9n + 19 = t , ta có :

( t - 1)( t + 1) = 1680

⇔ t² - 1 = 1680

⇔ t² - 41² = 0

⇔ ( t - 41)( t + 41) = 0

⇔ t = 41 hoặc t = - 41

+) Với : t = 41 , ta có :

n² - 9n + 19 = 41

⇔ n² - 9n - 22 = 0

⇔ n² + 2n - 11n - 22 = 0

⇔ n( n + 2) - 11( n + 2) = 0

⇔ ( n + 2)( n - 11) = 0

⇔ n = - 2 hoặc n = 11

+) Với : t = -41 ( giải tương tự )

@Giáo Viên Hoc24.vn

@Giáo Viên Hoc24h

@Giáo Viên

@giáo viên chuyên

@Akai Haruma

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

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

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

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 

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

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

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

\(x\left(2x-3\right)=0\)

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

bạn đăng tách ra nhé

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

\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)

b, sửa đề :  \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)

\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)

c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)

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

d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)

\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)

e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)

\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)

f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)

\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)

g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)

\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)

vô nghiệm 

d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)

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

\(=x^2-2x+1\)

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

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_#