a, Xét : x-4 = 0 => x= 4

            2x+1 = 0 => x= \(\frac{1}{2}\)

            x+3 = 0 => x = -3

            x + 9 = 0 => x = -9

Khi đó ta có bảng xét dấu : 

=> có 5 trường hợp:

TH1 : \(x\le-9\)

TH2 : \(-9\le x< -3\)

TH3 : \(-3\le x< \frac{1}{2}\)

TH4 : \(\frac{1}{2}\le x< 4\)

Do đó :

TH1 : \(x\le-9\)

Ta có :  /x-4/ = -(x-4) = 4 - x

            /2x+1/ = -(2x+1) = -2x -1

           /x+3/   = -(x + 3 ) = -x - 3

          /x-9/ = -(x-9) = -x + 9                  Thay vào đề bài ta có:

                                               3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5

                                    => 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5

                                    =>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5

                                   => -13 - 7x                                        = 5

                                             7x                                =     -13 - 5

                                                 7x =      -18

                                              x = \(\frac{-18}{7}\)( Ko TM)

Tương tự với 4 trường hợp còn lại.

a: =>2x>-6

hay x>-3

e: =>(5-x)/x<0

=>0<x<5

h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)

ai giúp vớ cần gấp

ĐK: \(\frac{2}{3}\le x\le\frac{3}{2}\)

(Vế phải và vế trái đều không âm nên có thể bình phương 2 vế theo một phương trình tương đương)

pt <=> \(x^2\left(3x-2\right)+\left(3-2x\right)+2\sqrt{x^2\left(3x-2\right)\left(3-2x\right)}=x^3+x^2+x+1\)

<=> \(3x^3-2x^2+3-2x+2\sqrt{x^2\left(3x-2\right)\left(3-2x\right)}-x^3-x^2-x-1=0\)

<=> \(2x^3-3x^2+2-3x+2\sqrt{x^2\left(3x-2\right)\left(3-2x\right)}=0\)

<=> \(x^2\left(2x-3\right)+\left(2-3x\right)+2\sqrt{x^2\left(3x-2\right)\left(3-2x\right)}=0\)

<=> \(-x^2\left(3-2x\right)-\left(3x-2\right)+2\sqrt{\left(3x-2\right).x^2\left(3-2x\right)}=0\)

<=> \(x^2\left(3-2x\right)+\left(3x-2\right)-2\sqrt{\left(3x-2\right).x^2\left(3-2x\right)}=0\)

<=> \(\left(\sqrt{x^2\left(3-2x\right)}-\sqrt{3x-2}\right)^2=0\)

<=> \(\sqrt{x^2\left(3-2x\right)}-\sqrt{3x-2}=0\)

<=> \(\sqrt{x^2\left(3-2x\right)}=\sqrt{3x-2}\)

<=> \(x^2\left(3-2x\right)=3x-2\)

<=> \(-2x^3+3x^2-3x+2=0\)

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

<=> x=1  (tm) 

ĐKXĐ: \(\frac{2}{3}\le x\le\frac{3}{2};x\in R\)

Pt cho tương đương: \(x\sqrt{3x-2}+\sqrt{3-2x}=\sqrt{\left(x+1\right)\left(x^2+1\right)}\)

Đặt \(\sqrt{3x-2}=a;\sqrt{3-2x}=b\left(a,b\ge0\right)\). Khi đó, ta được phương trình:

\(ax+b=\sqrt{\left(a^2+b^2\right)\left(x^2+1\right)}\Leftrightarrow a^2x^2+2abx+b^2=a^2x^2+b^2x^2+a^2+b^2\)

\(\Leftrightarrow2abx-b^2x^2-a^2=0\Leftrightarrow a^2-2abx+b^2x^2=0\)

\(\Leftrightarrow\left(a-bx\right)^2=0\Leftrightarrow a=bx\) hay \(\sqrt{3x-2}=x\sqrt{3-2x}\Leftrightarrow3x-2=3x^2-2x^3\)

\(\Leftrightarrow2x^3-3x^2+3x-2=0\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=9\)

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

Vậy PT cho có nghiệm duy nhất x=1.

Cái chỗ " 2(x-1)(x²+x+1) - 3x(x-1) = 9" bn sửa 9 thành 0 nhé, tại mik gõ vội :(

oki mơn bạn

\(\text{a) 2(x+3)-3(x-1)=2}\)

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

\(2x-3x=2-3-6\)

\(-x=-7\)

\(x=7\)

\(\text{b) 7-(x-2)=5(2x-3)}\)

\(7-x+2=10x-15\)

\(-x-10x=-15-2-7\)

\(-11x=-24\)

\(x=-24:\left(-11\right)\)

\(x=\frac{24}{11}\)

\(\text{c) 32-4(0,5y-5)=3y+2}\)

\(32-2y+20=3y+2\)

\(-2y-3y=2-20-32\)

\(-y=-50\)

\(y=50\)

\(\text{d) 3(x-1)-x=2x-3}\)

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

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

\(0=0\)( vô nghiệm )

a) 2(x + 3) - 3(x - 1) = 2

<=> 2x + 6 - 3x + 3 = 2

<=> -x + 9 = 2

<=> -x = -2 - 9

<=> -x = -7

<=> x = 7

b) 7 - (x - 2) = 5(2x - 3)

<=> 7 - x + 2 = 10x - 15

<=> 9 - x = 10x - 15

<=> 9 - x - 10 = -15

<=> 9 - 11x = -15

<=> -11x = -15 - 9

<=> -11x = -24

<=> x = 24/11

c) 32 - 4(0,5y - 5) = 3y + 2

<=> 32 - 2y + 20 = 3y + 2

<=> 52 - 2y = 3y + 2

<=> 52 - 2y - 3y = 2

<=> 52 - 5y = 2

<=> -5y = 2 - 52

<=> -5y = -50

<=> y = 10

\(e,x^2+2x-4=-12+3x+x^2\)

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

\(-x=-8\)

\(x=8\)

\(\text{g) x(x-1)-x(x+3)=15}\)

\(2x-x-2x-3x=15\)

\(-4x=15\)

\(x=15:\left(-4\right)\)

\(x=\frac{15}{-4}\)

\(\text{h) x(x-1)=x(x+3)}\)

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

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

\(-4x=0\)

\(\Rightarrow x=0\)

\(\text{i) x(2x-3)-x^2+2=x(x-5)-1}\)

\(3x-3x-x^2+2=2x-5x-1\)

\(3x-3x-2x+5x-x^2=-1-2\)

\(3x-x.x=-3\)

\(x.\left(3-1\right)=-3\)

\(\)\(x.2=-3\)

\(x=-3:2\)

\(x=\frac{-3}{2}\)

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

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

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

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

\(-x-x.x=0\)

\(-x.\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x=0\\1-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

\(\Rightarrow x\in\left\{0;1\right\}\)

chúc bạn học tốt

ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

                          \(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)

                      \(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............

a) Ta có:

f(0) = -2.0³ + 3.0² - 0 + 5 = 0 + 0 - 0 + 5 = 5

g(-1) = 2.(-1)³ - 2.(-1)² + (-1) - 9 = -2 - 2 - 1 - 9 = -14

b) f(x) + g(x) = (-2x³ + 3x² - x + 5) + (2x³ - 2x² + x - 9)

                   = -2x³ + 3x² - x + 5 + 2x³ - 2x² + x - 9

                  = (-2x³ + 2x³) + (3x² - 2x²) - (x - x) + (5 - 9)

                 = x² - 4

f(x) - g(x) = (-2x³ + 3x² - x + 5) - (2x³ - 2x² + x - 9)

               = -2x³ + 3x² - x + 5 - 2x³ + 2x² - x + 9

              = -(2x³ + 2x³) + (3x² + 2x²) - (x + x) + (5 + 9)

             = -4x³ + 5x² - 2x + 14

\(\Rightarrow4x+8-3+6x=7-2x-2\)

\(\Rightarrow4x+6x+2x=-8+3-2\)

\(\Rightarrow12x=-7\)

\(\Rightarrow x=\frac{-7}{12}\)

Vậy \(x=\frac{-7}{12}.\)

bạn hoàng thị ngọc ánh làm sai rồi

\(4.\left(x+2\right)-3\left(1-2x\right)=7-2\left(x+1\right)\)

\(\Rightarrow4x+8-3-6x=7-2x-2\)

\(\Rightarrow4x+2x-6x+8-3=5\)

\(\Rightarrow6x-6x+8=5+3\)

\(\Rightarrow6x-6x=8-8\)

\(\Rightarrow6x-6x=0\)

2x-3=x+1/2

a,2x-3=x+1/2                       b,4x-(x+1/2)=2x+(1/2-5)                           c,2/3-1/3(x-2/3)-1/2(2x+1)=5

2x-x =1/2+3                           4x-x-1/2=2x+1/2-5                             d,(x+1/2).(x-3/4)=0

x=7/2                                4x-x-2x  =1/2-5+1/2                                 \(\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

                                            x=-4

e,(2x-1)(3x+1/5)=0

\(\orbr{\begin{cases}2x-1=0\\3x+\frac{1}{5}=0\end{cases}}\orbr{\begin{cases}2x=1\\3x=\frac{1}{5}\end{cases}}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{15}\end{cases}}\)

f, 4x²-2x=0

Các câu mk chưa làm thì bạn cứ chờ để mk suy nghĩ.

a) = (x + 3)² - y² = (x + 3 - y)(x + 3 + y)

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

c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)

d) Nhầm đề. tui sửa lại x³ + y³ + 2x² - 2xy + 2y²

= x³ + y³ + 2(x² - xy + y²) = (x + y)(x² - xy + y²) + 2(x² - xy + y²) = (x² - xy + y²)(x + y + 2)

e) = x⁴ - x³ - x³ + x² - x² + x + x - 1 = x³(x - 1) - x²(x - 1) - x(x - 1) + x - 1 = (x - 1)(x³ - x² - x + 1) = (x - 1)(x - 1)(x² - 1) = (x - 1)³(x + 1)

f) = x³ - 3x² - x² + 3x + 9x - 27 = x²(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x² - x + 9)

g) chắc là 3xyz 

= x²y + xy² + y²z + yz² + x²z + xz² + 3xyz = x²y + xy² + xyz + y²z + yz² + xyz + x²z + xz² + xyz = (x + y + z)(xy + yz + xz)

h) = 2³ -(3x)³ = (2 - 3x)(4 + 6x + 9x²)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x³ - y³)(x³ + y³) = (x - y)(x² + xy +y2)(x + y)(x² - xy +y2).