( 7 + x ) * 4 + x = 108

=> 28 + 4x + x = 108

=> 28 + 5x = 108

=> 5x = 80

=> x = 16 (tm)

(7 + x ) * 4 + x = 108

7 * 4 + x * 4 + x = 108

28 + x * 5 = 108

x * 5  = 108 -  28

x * 5 = 80

x = 80 : 5

x = 16

( 7 + x ) x 4 + x = 108

7 x 4 + x x 4 + x = 108

28 + x x 5 = 108

x x 5 = 108 - 28

x x 5 = 80

x = 80 : 5

x = 16

(7+X)x4+X=108

=> 28+4X+X=108

=> 5X=108-28

=> 5X =80

=> X=16 

\(\dfrac{4}{5}\times\dfrac{3}{7}\times\dfrac{4}{5}\times\dfrac{6}{7}\times\dfrac{4}{14}=\dfrac{4}{5}\times\dfrac{3}{7}+\dfrac{4}{5}\times\dfrac{6}{7}-\dfrac{4}{5}\times\dfrac{2}{7}=\dfrac{4}{5}\times\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right)=\dfrac{4}{5}\)

=15,8 x (108 - 7)

=15,8 x 101

=

`15,8xx108-15,8xx7-15,8`

`=15,8xx(108-7-1)`

`=15,8xx100`

`=1580`

1580 nha

11340

14472

ghi rõ đề bài ra nào

a) 4 giờ 22 phút x 4 + 5 giờ 38 phút x 4

= ( 4 giờ 22 phút + 5 giờ 38 phút ) x 4

= 10 giờ x 4

=40 giờ

b) 7 phút 30 giây x 7 - 6 phút 30 giây x 7

= ( 7 phút 30 giây - 6 phút 30 giây ) x 7

= 1 phút x 7

= 7 phút

Kết bn với mik nha!

\(A\left(x\right)+B\left(x\right)-C\left(x\right)\)

\(=\left(-7+2x^2+x^4+3x^5-x^3\right)+\left(-x+x^4+2x^3-7\right)-\left(2x-x^4-3x^3\right)\)

\(=3x^5+3x^4+4x^3+2x^2-3x-14\)

d: \(x\left(x+1\right)\left(x^2+x+1\right)=42\left(1\right)\)

=>\(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(a=x^2+x\)

Phương trình (1) sẽ trở thành \(a\left(a+1\right)=42\)

=>\(a^2+a-42=0\)

=>(a+7)(a-6)=0

=>\(\left(x^2+x+7\right)\left(x^2+x-6\right)=0\)

mà \(x^2+x+7=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}>0\forall x\)

nên \(x^2+x-6=0\)

=>(x+3)(x-2)=0

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

e: \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\left(2\right)\)

=>\(\left(x-1\right)\left(x+5\right)\left(x-3\right)\left(x+7\right)-297=0\)

=>\(\left(x^2+4x-5\right)\left(x^2+4x-21\right)-297=0\)

Đặt \(b=x^2+4x\)

Phương trình (2) sẽ trở thành \(\left(b-5\right)\left(b-21\right)-297=0\)

=>\(b^2-26b+105-297=0\)

=>\(b^2-26b-192=0\)

=>(b-32)(b+6)=0

=>\(\left(x^2+4x-32\right)\left(x^2+4x+6\right)=0\)

mà \(x^2+4x+6=\left(x+2\right)^2+2>0\forall x\)

nên \(x^2+4x-32=0\)

=>(x+8)(x-4)=0

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

f: \(x^4-2x^2-144x-1295=0\)

=>\(x^4-7x^3+7x^3-49x^2+47x^2-329x+185x-1295=0\)

=>\(\left(x-7\right)\cdot\left(x^3+7x^2+47x+185\right)=0\)

=>\(\left(x-7\right)\left(x+5\right)\left(x^2+2x+37\right)=0\)

mà \(x^2+2x+37=\left(x+1\right)^2+36>0\forall x\)

nên (x-7)(x+5)=0

=>\(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

e: =>x(x^3-4x^2-8x+8)=0

=>x[(x^3+8)-4x(x+2)]=0

=>x(x+2)(x^2-2x+4-4x)=0

=>x(x+2)(x^2-6x+4)=0

=>\(x\in\left\{0;-2;3+\sqrt{5};3-\sqrt{5}\right\}\)

g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0

=>(2x+5)(x^3-3x^2+3x-1)=0

=>(2x+5)(x-1)^3=0

=>x=1 hoặc x=-5/2

h: =>(x^2+8x+7)(x^2+8x+15)+15=0

=>(x^2+8x)^2+22(x^2+8x)+120=0

=>(x^2+8x+10)(x^2+8x+12)=0

=>(x^2+8x+10)(x+2)(x+6)=0

=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)