x *6+ ( 1 + 2 + 3 + 4 + 5 ) = 45

x*6=45-15=30

x=30:6=5

ủng hộ nha các bạn

khong phai toan lop 3

x+1 + x+2 + x+3 + x +4 + x+5 = 45

x + x + x + x + x = 45 -1 -2 -3 -4 -5

5x = 30

x =6

(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45 

(x + x + x + x + x) + ( 1 + 2 + 3 + 4 + 5)  = 45

\(x\times5\)+ 15 = 45

\(x\times5\)       = 45 - 15

\(x\times5\)        = 30

x                      = 30 : 5

x                      = 6.

Hok tốt !

b: ĐKXĐ: x∉{-3;1;-1}

Ta có: \(\frac{1}{x^2+2x-3}=\frac{1}{\left(x+1\right)^2}+\frac{1}{48}\)

=>\(\frac{1}{\left(x+3\right)\left(x-1\right)}-\frac{1}{\left(x+1\right)^2}=\frac{1}{48}\)

=>\(\frac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)

=>\(\frac{x^2+2x+1-\left(x^2+2x-3\right)}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)

=>\(\frac{4}{\left(x+1\right)^2\cdot\left(x-1\right)\left(x+3\right)}=\frac{1}{48}\)

=>\(\left(x+1\right)^2\cdot\left(x^2+2x-3\right)=192\)

=>\(\left(x^2+2x+1\right)\left(x^2+2x-3\right)=192\)

=>\(\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)

=>\(\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)

mà \(x^2+2x+13=\left(x+1\right)^2+12\ge12>0\forall x\)

nên \(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)

5x+15=45

5x=35

x=7

(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45

( x + x + x + x + x ) + ( 1 + 2 + 3 + 4 + 5 ) = 45

5x + 15 = 45

5x = 45 - 15

5x = 30

x = 30 : 5

x = 6

1) x = 4/5 - 1/3

x = 7/15

2) 5/3.x=1/21

x=1/35

3) -12/13.x = 1/13

x=-1/12

7) th1: x-1=2/3

x = 5/3

Th2: x - 1 = -2/3

x=1/3

(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45

<=> (x+x+x+x+x)+(1+2+3+4+5)=45

<=> 5x+15=45

<=> 5x=30

<=> x=6

(x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 45

x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 45

(x + x + x + x + x) + (1 + 2 + 3 + 4 + 5) = 45

5x + 15 = 45

5x = 45 - 15

5x = 30

x = 6