tìm y nữa 

mình viết thiếu

5-x-16=40+x

-x-x=40+16-5

(-x).2=51

-x=51/2=>x=51/2

1.x=40

2.x=10

3.x=

15-x = 4x -5 

15 + 5 = 4x + x 

5x = 20

x= 4

x - 15 = 6 + 4x 

-15 - 6 = 4x -x 

3x = -21 

x= -7

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

Nhìu quá bạn ơi

nhiều vậy ai mà làm dc bạn ơi

làm hai câu ba câu cũng được

(Phần a mình lấy vế phải bằng 0 nha ^^)

a,

\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)+7=0\\ \Leftrightarrow25x^2-10x+1-\left(25x^2-16\right)+7=0\\ \Leftrightarrow25x^2-10x+1-25x^2+16+7=0\\ \Leftrightarrow-10x+24=0\\ \Leftrightarrow x=2,4\)

b,

\(5x^2+4xy+4y^2+4x+1=0\left(1\right)\\ \Leftrightarrow4x^2+4x+1+x^2+4xy+4y^2=0\\ \Leftrightarrow\left(2x+1\right)^2+\left(x+2y\right)^2=0\left(1a\right)\)

Do \(VT\ge0\) với \(\forall x,y\in R\) nên:

\(\left(1a\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

c,

\(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\\ \Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)-6x^2-21=0\\ \Leftrightarrow x^3+12x+8-x^3+x-21=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow x=1\)

Chúc bạn học tốt nha.

\(b)5x^2 + 4xy + 4y^2 + 4x + 1 = 0\)

\(\Leftrightarrow\) \(4x^2 + 4x + 1 + x^2 + 4xy + 4y^2 = 0\)

\(\Leftrightarrow\)\((2x + 1)^2 + (x + 2y)^2 = 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

\(c)(x+2)^3-x(x-1)(x+1)=6x^2+21\)

\(\Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)=6x^2+21\\ \Leftrightarrow13x+8=21\\ \Leftrightarrow13x=21-8\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)

x^2=16

Suy ra x^2=4^2

Vậy x^2=4^2

[1/(x+2)-1/(x+5)]+[1/(x+5)-1/(x+10)]+[1/(x+10)-1/(x+17)]=x/15.[1/(x+2)-1/(x+17)]
1/(x+2)-1/(x+17)=x/15.[1/(x+2)-1/(x+17)]
1=x/15
x=15

sdsdxc

1,-12 + x = 5x - 20

-12 + 20 = 5x - x

8 = 4x

x = 2

2, 7x - 4 = 20 + 3x

7x - 3x = 20 + 4

4x = 24

x  = 6

3 , 5x - 7 = -21 - 2x

5x + 2x  = -21 + 7

7x = -14

x = -14 : 7 = -2

4 , x + 15  = 20 - 4x

x + 4x = 20 - 15

5x = 5

x = 5 : 5 = 1

5, 17 - x = 7 - 6x

-x + 6x = 7 - 17

5x = -10

x = -10 : 5 = -2