20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

Mấy cái này chuyển vế đổi dấu là xong í mà :3

1,

16-8x=0

=>16=8x

=>x=16/8=2

2, 

7x+14=0

=>7x=-14

=>x=-2

3,

5-2x=0

=>5=2x

=>x=5/2

Mk làm 3 cau làm mẫu thôi

Lúc đăng đừng đăng như v :>

chi ra khỏi ngt nản

từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại

1, 16 - 8x = 0

<=>-8x = 16

<=> x = -2

Vậy_

2, 7x + 14 = 0

<=> 7x = -14

<=> x = -2

3, 5 - 2x = 0

<=> - 2x = -5

<=> x =\(\frac{5}{2}\)

Vậy_

4, 3x - 5 = 7

<=> 3x = 7 + 5

<=> 3x = 12

<=> x = 4

Vậy...

5, 8 - 3x = 6

<=> - 3x = 6 - 8

<=> -3x = - 2

<=> x =\(\frac{2}{3}\)

Vậy......

Trả lời

Bài 1:

a)(45-25).(-11)+29.(-3-17)

=20.(-11)+29.(-20)

=20.(-11)+(-29).20

=20.[(-11)+(-29)]

=20.(-30)

=-600

b)(36-6).(-5)+21.(-17-3)

=30.(-5)+21.(-20)

=(-150)+(-420)

=-570

trả lời:

a,(45-25).(-11)+29.(-3-17)

  =20.(-11)+29.(-20)

  =20.(-11)+(-29).20

  =20.[(-11)+(-29).20

  =20.(-30)

  =-600

học tốt

I

phần cuối mk chụp ko đc hết . chỗ cuối là bằng \(\frac{-5}{-3}\)=\(\frac{5}{3}\)

Bài 1:

a) Ta có: 7x+12=0

\(\Leftrightarrow7x=-12\)

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

Vậy: \(x=-\frac{12}{7}\)

b) Ta có: 5x-2=0

\(\Leftrightarrow5x=2\)

hay \(x=\frac{2}{5}\)

Vậy: \(x=\frac{2}{5}\)

c) Ta có: 12-6x=0

\(\Leftrightarrow6x=12\)

hay x=2

Vậy: x=2

d) Ta có: -2x+14=0

⇔-2x=-14

hay x=7

Vậy: x=7

Bài 2:

a) Ta có: 3x+1=7x-11

⇔3x+1-7x+11=0

⇔-4x+12=0

⇔-4x=-12

hay x=3

Vậy: x=3

b) Ta có: 2x+x+12=0

⇔3x+12=0

⇔3x=-12

hay x=-4

Vậy: x=-4

c) Ta có: x-5=3-x

⇔x-5-3+x=0

⇔2x-8=0

⇔2x=8

hay x=4

Vậy: x=4

d) Ta có: 7-3x=9-x

⇔7-3x-9+x=0

⇔-2x-2=0

⇔-2x=2

hay x=-1

Vậy: x=-1

e) Ta có: 5-3x=6x+7

⇔5-3x-6x-7=0

⇔-9x-2=0

⇔-9x=2

hay \(x=\frac{-2}{9}\)

Vậy: \(x=\frac{-2}{9}\)

f) Ta có: 11-2x=x-1

⇔11-2x-x+1=0

⇔12-3x=0

⇔3x=12

hay x=4

Vậy: x=4

g) Ta có: 15-8x=9-5

⇔15-8x=4

⇔8x=11

hay \(x=\frac{11}{8}\)

Vậy: \(x=\frac{11}{8}\)

Bài 3:

a) Ta có: 0,25x+1,5=0

⇔0,25x=-1,5

hay x=-6

Vậy: x=-6

b) Ta có: 6,36-5,2x=0

⇔5,2x=6,36

hay \(x=\frac{159}{130}\)

Vậy: \(x=\frac{159}{130}\)

\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)

\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)

.......................................................................................

\(x^3-8x^2-8x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)

......................................................................................

cảm ơn nha 

Sửa đề \(\left(8x-11\right)^3+\left(7x-12\right)^3+\left(23-15x\right)^3=0\)

Đặt \(8x-11=a\)

\(7x-12=b\)

\(23-15x=c\)

=> a+b+c=8x-11+7x-12+23-15x=0

Có \(a^3+b^3+c^3-3abc\)

= \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

=\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)

=0 (do a+b+c=0)

=> \(a^3+b^3+c^3=3abc\)

<=> \(0=3\left(8x-11\right)\left(7x-12\right)\left(23-15x\right)\)

=> \(\left[{}\begin{matrix}x=\frac{11}{8}\\x=\frac{12}{7}\\x=\frac{23}{15}\end{matrix}\right.\)

a. 7x+12= 0 \(\Leftrightarrow7x=-12\Leftrightarrow x=-\frac{12}{7}\)

b.-2x+14=0 \(\Leftrightarrow-2x=-14\Leftrightarrow x=7\)

c. 3x+1=7x-11 \(\Leftrightarrow3x-7x=-11-1\Leftrightarrow-4x=-12\Leftrightarrow x=3\)

d.2x+x+12=0 \(\Leftrightarrow2x+x=-12\Leftrightarrow3x=-12\Leftrightarrow x=-4\)

e.x-5=3-x \(\Leftrightarrow x+x=3+5\Leftrightarrow2x=8\Leftrightarrow x=4\)

f. 7-3x=9-x \(\Leftrightarrow-3x+x=9-7\Leftrightarrow-2x=2\Leftrightarrow x=-1\)

g. 8-3x=6x+7 \(\Leftrightarrow-3x-6x=7-8\Leftrightarrow-9x=-1\Leftrightarrow x=\frac{1}{9}\)

h. 11-2x=x-1\(\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=4\)

k. 15-8x=9-5x \(\Leftrightarrow-8x+5x=9-15\Leftrightarrow-3x=-6\Leftrightarrow x=2\)

l. 3+2x=5+2 \(\Leftrightarrow2x=5+2-3\Leftrightarrow2x=4\Leftrightarrow x=2\)

a, \(7x+12=0\)

\(\Leftrightarrow7x=-12\)

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

b, \(-2x+14=0\)

\(\Leftrightarrow-2x=-14\)

\(\Leftrightarrow x=7\)

c,\(3x+1=7x-11\)

\(\Leftrightarrow7x-3x=1+11\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=3\)

d,\(2x+x+12=0\)

\(\Leftrightarrow3x=-12\)

\(\Leftrightarrow x=-4\)

e,\(x-5=3-x\Leftrightarrow x+x=3+5\Leftrightarrow2x=8\Leftrightarrow x=4\)

f, \(7-3x=9-x\Leftrightarrow-3x+x=9-7\Leftrightarrow-2x=2\Leftrightarrow x=-1\)

g,\(8-3x=6x+7\Leftrightarrow6x+3x=8-7\Leftrightarrow9x=1\Leftrightarrow x=\frac{1}{9}\)

h,\(11-2x=x-1\Leftrightarrow x+2x=11+1\Leftrightarrow3x=12\Leftrightarrow x=4\)

k,\(15-8x=9-5x\Leftrightarrow-5x+8x=15-9\Leftrightarrow3x=6\Leftrightarrow x=2\)

l,\(3+2x=5+2\)

\(\Leftrightarrow2x=5+2-3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)