a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

[9x3(x2-1)-6x2(x2-1)2+12x(x2-1)]:3x(x2-1)

Gọi x2-1 là y tá đc

(9x3y-6x2y2+12xy):3xy

=3x2-2xy+4

=3x2-2x(x2-1)+4

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)

Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)

                    \(x=0\)

Lập bảng xét dấu :

     \(x\)                           \(0\)                   \(\dfrac{9}{16}\)

\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\)         \(-\)       \(0\)           \(-\)       \(0\)        \(+\)

      \(\left|x\right|\)              \(-\)       \(0\)           \(+\)       \(0\)        \(+\)

TH1 : \(x< 0\)

\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)

\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)

\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))

TH2 : \(0\le x\le\dfrac{9}{16}\)

\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)

\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))

TH3 : \(x>\dfrac{9}{16}\)

\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)

\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))

Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)

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

b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)

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

d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)

`@` `\text {Ans}`

`\downarrow`

b) ( x+ 2) ( 3x-2) - (3x-1) ( x-5) = 11

3x^2 +6x - 2x -4 - 3x^2 +x + 15x -5 =11

20x = 11 +4+5

20x = 20

x=1

c) 2(2x+ 1) ( 8x-3) + ( 3-4x) ( 8x-7) = 6x + 73

(4x + 2)(8x-3) + 24x - 32x^2 -3 +4x = 6x +73

32x^2 + 16x - 12x  + 24x - 32x^2 +4x -6x = 73 +6 +3

26x = 82

x= 41/13

a. \(\left(x+2\right)\left(3x-2\right)-\left(3x-1\right)\left(x-5\right)=11\)

\(\Rightarrow3x^2-2x+6x-4-3x^2+15x+x-5=11\)

\(\Rightarrow20x-9=11\)

\(\Rightarrow x=1\)

Vậy..................

b. \(2\left(2x+1\right)\left(8x-3\right)+\left(3-4x\right)\left(8x-7\right)=6x+73\)

\(\Rightarrow\left(4x+2\right)\left(8x-3\right)+\left(3-4x\right)\left(8x-7\right)=6x+73\)

\(\Rightarrow32x^2-12x+16x-6+24x-21-32x^2+28x-6x=73\)

\(\Rightarrow50x-27=73\)

\(\Rightarrow x=100\)

Vậy..............