x(x-5).(x+5)-(x+2).(x^2-2x+4)=17

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8=17\)

\(\Leftrightarrow-25x=17+8\)

\(\Leftrightarrow-25x=25\)

\(\Leftrightarrow x=-1\)

#)Giải :

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(\Rightarrow x^3-25-\left(x^3+8\right)=17\)

\(\Rightarrow x^3-25x-x^3-8=17\)

\(\Rightarrow-25x=25\Rightarrow x=-1\)

Vậy x = -1

x.(x - 5).(x + 5) - (x + 2).(x² - 2x + 4) = 17

<=> -25x - 8 = 17

<=> -17x = 17

<=> x = -1

=> x = -1

x(x - 5)(x + 5) - (x + 2)(x² - 2x + 4) = 17

=> x(x² - 25) - (x³ + 2³) = 17

=> x³ - 25x - x³ - 8 = 17

=> 25x - 8 = 17

=> 25x = 17 + 8

=> 25x = 25

=> x = 1

#)Giải :

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Rightarrow x^3-25x-\left(x^3+8\right)=17\)

\(\Rightarrow x^3-25x-x^3-8=17\)

\(\Rightarrow-25x=25\Leftrightarrow x=-1\)

#)Giải :

\(2x^3-3x^2+x+m=\left(x+2\right)\left(2x^2-7x+15\right)\)

\(\Leftrightarrow2x^3-3x^2+x+m=2x^3-7x^2+15x+4x^2-14x+30\)

\(\Leftrightarrow-3x^2+x+m=-3x^2+x+30\)

\(\Leftrightarrow m=30\)

Sorry . I am class 7a

Thay x=0 vào y=x-3, ta được:

y=0-3=-3

Thay x=0 và y=-3 vào (d), ta được:

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

=>m+1=-3

=>m=-4

( x - 1)³ - (x + 3) . (x² - 3x + 9) + 3 .  (x + 2) - (x - 2)  = 2

=>x³-3x².(-1)+3x.(-1)²-(-1)³-x(x²-3x+9)-3(x²-3x+9)+3x+6-x+2=2

x³+3x²+3x+1-x³+3x²-9x-3x²+9x-27+3x+6-x+2=2

(x³-x³)+(3x²+3x²-3x²)+(3x-9x+9x+3x-x)+(1-27+6+2)=2

3x²-5x-18=2

x(3x-5)=20

Thử lần lượt nha bạn

Bài 2

(x+y+z)²-2(x+y+z)(x+y)+(x+y)²

=(x+y+z)²-2x²-4xy-2xz-2yz+x²+2.xy+y²

=z²+(y+x)2z+y²+2xy+x²-2x²-4xy-2z(x+y)+x²+2xy+y²

=z²+(x+y)2z-2z(x+y)+(y²+y²)+(2xy+2xy-4xy)+(x²-2x²+x²)

=z²+2y²

\(\dfrac{5}{2}\times\dfrac{1}{3}+\dfrac{1}{4}\\ =\dfrac{5}{6}+\dfrac{1}{4}\\ =\dfrac{10}{12}+\dfrac{3}{12}\\ =\dfrac{13}{12}\)

= 5/6 + 1/4

= 10/12 + 3/12

= 13/12

Vậy giá trị biểu thức là 13/12

Nhân chia trc cộng trừ sau:

= 5/6 + 1/4

= 10/12 + 3/12

= 13/12

\(x-1+x-4=3x\)

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

\(\Leftrightarrow0+\left(-5\right)=3x\)

\(\Leftrightarrow3x=-5\)

\(\Leftrightarrow x=-\frac{5}{3}\)

Vậy : \(x=-\frac{5}{3}\)

Mình sửa lại nhé, chưa đọc kĩ đề :

\(x-1+x-4=3x\)

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

\(\Leftrightarrow2x-5=3x\)

\(\Leftrightarrow3x-2x=-5\)

\(\Leftrightarrow x=-5\)

Vậy \(x=-5\)