a) 3 -4u + 24 + 6u = u + 27 +3u

=> -21 +2u = 27 +4 u

=> -2u = 48

=> u = -24

b) -6(1.5 -2x ) = 3( -15 +2x )

=> -9 +12x = -30 + 6x

=> 6x = -21

=> x = \(\frac{-7}{3}\)

c ) 0.1 -2( 0.5t - 0.1 ) = 2( t-2.5 ) -0.7

=>0.1 -1t+ 0.2 = 2t-5-0.7

=>0.1+5.7 = 1t +2t

=> 5.8 = 3t

=> t = \(\frac{5.8}{3}\)

câu này là at hay on vậy

at

bài lớp 7 gì căng vậy

Câu 1: 

Chu vi của tem là 50x1/5=10(cm)

Câu 2: B

a, 3x -2 = 2x - 3 

=> 3x - 2x = 2 - 3 

=> x= - 1

b, là tương tự câu a 

các câu sau bạn nhân phá ra mà giải nhé

a, 3x - 2 = 2x - 3

3x - 2x = -3 + 2

x = -1

b, 3 - 4u + 24 + 6u = u + 27 + 3u

-4u + 6u - u - 3u = 27 - 3 - 24

-2u = 0

u = 0 : (-2)

u = 0

c, 5 - (x - 6) = 4(3 - 2x)

5 - x + 6 = 12 - 8x

-x + 8x = 12 - 5 - 6

7x = 1

x = 1/7

d, -6(1,5 - 2x) = 3(-15 + 2x)

-9 + 12x = -45 + 6x

12x - 6x = -45 + 9

6x = -36

x = (-36) : 6

x = -6

e, 0,1 - 2(0,5 - 0,1) = 2(t - 2,5) - 0,7

0,1 - 1 + 0,2 = 2t - 5 - 0,7

-2t = -5 - 0,7 - 0,1 + 1 - 0,2

-2t = -5

t = -5/-2

t = 5/2

a) 3x - 2 = 2x - 3

⇔ 3x - 2x = -3 + 2

⇔ x          = -1

Vậy phương trình có nghiệm duy nhất x = -1.

b) 3 - 4u + 24 + 6u = u + 27 + 3u

⇔ 2u + 27           = 4u + 27

⇔ 2u - 4u            = 27 - 27

⇔ -2u                  = 0

⇔ u                     = 0

Vậy phương trình có nghiệm duy nhất u = 0.

c) 5 - (x - 6) = 4(3 - 2x)

⇔ 5 - x + 6 = 12 - 8x

⇔ -x + 11   = 12 - 8x

⇔ -x + 8x   = 12 - 11

⇔ 7x          = 1

⇔ x            = 17

Vậy phương trình có nghiệm duy nhất x = 17.

d) -6(1,5 - 2x) = 3(-15 + 2x)

⇔ -9 + 12x      = -45 + 6x

⇔ 12x - 6x      = -45 + 9

⇔ 6x               = -36

⇔ x                 = -6

Vậy phương trình có nghiệm duy nhất x = -6

Bài 1:

a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{2-u}{u+2}\)(1)

Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)

\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)

\(=\frac{-\left(u-2\right)}{u+2}\)

\(=\frac{2-u}{u+2}\)(2)

Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)

b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)

\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)

\(=v+3=VP\)(đpcm)

Bài 2:

a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)

\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)

\(\Leftrightarrow M=2x^2-3x+2x-3\)

hay \(M=2x^2-x-3\)

Vậy: \(M=2x^2-x-3\)

b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)

\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)

\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{M}{\left(x-2\right)^2}\)

\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)

\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)

\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)

\(\Leftrightarrow M=2x^2-4x-x+2\)

hay \(M=2x^2-5x+2\)

Vậy: \(M=2x^2-5x+2\)

Bài 3:

a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)

\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)

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

hay \(N=x^2+3x+2\)

Vậy: \(N=x^2+3x+2\)

n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)

\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)

\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)

hay \(N=\frac{2x-6}{x+3}\)

Vậy: \(N=\frac{2x-6}{x+3}\)