Bài 3:

a: =>3x^2-6x-x-3x^2=14

=>-7x=14

=>x=-2

b: \(\Leftrightarrow2x^2+10x-x-5-2x^2-9x-x-4.5=3.5\)

=>-x-9,5=3,5

=>-x=12

=>x=-12

c: =>\(3x-3x^2+9x=36\)

=>-3x^2+12x-36=0

=>x^2-6x+12=0(loại)

d: \(\Leftrightarrow3x^2-3x+x-1+4x-3x^2=5\)

=>2x=6

=>x=3

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

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

\(\Leftrightarrow3=3\)( Luôn đúng với mọi x )

Vậy phương trình nghiệm đúng với mọi x

b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)

\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)

\(\Leftrightarrow-3x^3+6x^2-3x-24=12x+12\)

\(\Leftrightarrow-3x^3+6x^2-3x-24-12x-12=0\)

\(\Leftrightarrow-3x^3+6x^2-15x-36=0\)

Đến đây xem lại đề bạn nhớ :D Tìm thì tìm được nhưng thấy nó sai sai kiểu gì í

c) \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)

\(\Leftrightarrow3x\left(x-2\right)+1\left(x-2\right)=2\left(-3x-5\right)-x\left(-3x-5\right)\)

\(\Leftrightarrow3x^2-6x+x-2=-6x-10+3x^2+5x\)

\(\Leftrightarrow3x^2-6x+x+6x-3x^2-5x=-10+2\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=2\)

d) \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x\left(x+5\right)+3\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x^2+5x+3x+15-x^2-7x=2x+8\)

\(\Leftrightarrow x^2+5x+3x-x^2-7x-2x=8-15\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

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

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

\(\Leftrightarrow-2x=0\Leftrightarrow x=0\)

b, \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)

\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)

\(\Leftrightarrow-3x-24+6x^2-3x^3=12x+12\)

\(\Leftrightarrow-15x-36+6x^2-3x^3=0\)

Lớp 8 chưa hc vô tỉ đâu ... vô nghiệm 

c, \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)

\(\Leftrightarrow3x^2-5x-2=-x-10+3x^2\)

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

d, \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x^2+8x+15-x^2-7x=2x+8\)

\(\Leftrightarrow x+15=2x+8\Leftrightarrow-x+7=0\Leftrightarrow x=7\)

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

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

a. 3x² - 2x - 1 = 0

<=> 3x² - 3x + x - 1 = 0

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

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

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)

<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)

<=> 20x + 20 + 24x + 36 = 45

<=> 44x = -11

<=> x = \(-\dfrac{1}{4}\)

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

    \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)

       \(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)

Bài 5 :

a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

=> \(36x+3=0\)

=> \(x=-\frac{1}{12}\)

Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)

b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)

=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)

=> \(35x-5+60x-96+6x=0\)

=> \(101x-101=0\)

=> \(x=1\)

Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)

c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

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

=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)

=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)

=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

=> \(-64x+123=0\)

=> \(x=\frac{123}{64}\)

Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)

1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

\(y'=\dfrac{3}{\left(-x+1\right)^2}\)

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)

6.

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