Ta có :(-37)-|7-x|=-127

suy ra | 7-x | =90

suy ra 7-x=90 hoặc 7-x=-90

Nếu 7-x=90 suy ra x=-83

Nếu 7-x=-90 suy ra x =97

Vậy x=-83 hoac x=97

1: 7-x=8+(-7)

=>7-x=8-7=1

=>x=7-1=6

2: \(x-8=\left(-3\right)-8\)

=>x-8=-11

=>\(x=-11+8=-3\)

3: \(2-x=10-9+23\)

=>\(2-x=33-9=24\)

=>x=2-24=-22

4: \(-2-x=15\)

=>\(x=-2-15=-17\)

5: \(-7+x-8=-3-1+13\)

=>x-14=13-4=9

=>x=9+14=23

6: 100-x+7=-x+3

=>107-x=3-x

=>107=3(vô lý)

7: \(23+x=8-2x\)

=>\(x+2x=8-23\)

=>3x=-15

=>x=-15/3=-5

a) (-x+31) -39= -69+11

 (-x+31) -39= -58

-x+31=-58+39

-x+31=-19

-x=-19-31

-x=-50

x=50

b)-129-(35-x) =55

35-x=-129-55

35-x=-184

x=35-(-184)

x=219

c) (-37)-l 7-yl =-127

l 7-yl =(-37)-(-127)

l 7-yl =90

=>7-y=90              hoặc 7-y=-90

       y=7-90                       y=7-(-90) 

       y=-83                         y=97

d) (2y-5)² =9

(2y-5)² =3²

=>2y-5=3

2y=3+5

2y=8

y=8:2

y=4

e)(1-3x)³= -8

(1-3x)³= (-2)³

1-3x=-2

3x=1-(-2)

3x=3

x=3:3

x=1

**** nhé

a) (-x + 31) - 39 = -69 + 11

=> - x + 31 - 39 = -58

=> - x = - 58 - 31 + 39

=> - x = - 50

=> x = 50

vậy_

a)

x − 5 12 − 4 9 = − 13 18 x = − 13 18 + 5 12 + 4 9 x = − 26 36 + 15 36 + 16 36 x = 5 36

b)

2 − x + 3 7 = 9 − 21 2 − x − 3 7 = − 3 7 − x = − 3 7 + 3 7 − 2 − x = − 2 ⇒ x = 2

\(C^n_n+C^{n-1}_n+C^{n-2}_n=37\)

\(\Leftrightarrow1+\dfrac{n!}{\left(n-1\right)!}+\dfrac{n!}{\left(n-2\right)!2!}=37\)

\(\Leftrightarrow1+n+\dfrac{n\left(n-1\right)}{2}=37\)

\(\Rightarrow n=8\)

\(P=\left(2+5x\right)\left(1-\dfrac{x}{2}\right)^8=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{x}{2}\right)^k\right)\)

\(=\left(2+5x\right).\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5x\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\)

\(=2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)+5\)\(\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\)

Số hạng chứa \(x^3\) trong \(2.\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^k\right)\) là \(2C^3_8.\left(-\dfrac{1}{2}\right)^3x^3\)

Số hạng chứa \(x^3\) trong \(5\left(\sum\limits^8_{k=0}.C_8^k.\left(-\dfrac{1}{2}\right)^k.x^{k+1}\right)\) là \(5C^2_8.\left(-\dfrac{1}{2}\right)^2x^3\)

Vậy số hạng chứa x³ trong P là:\(\left[2.C^3_8\left(-\dfrac{1}{2}\right)^3+5C^2_8\left(-\dfrac{1}{2}\right)^2\right]x^3\)

sao gữi 2 cái zữ z bn ???

-37 - | 7-x | = -127

| 7-x | = -37 - (-127)

| 7-x | = 90

1) 7-x = 90

x = 90 + 7

x = 97

2) 7-x = -90

x = -90+7

x = -83

x = 97; -83

\(-37-\left|7-x\right|=-127\)

th1 : \(x\le7\) thì \(-37-\left|7-x\right|=-127\)

\(\Leftrightarrow\) \(-37-\left(7-x\right)=-127\) \(\Leftrightarrow\) \(-37-7+x=-127\)

\(\Leftrightarrow\) \(x=-127+37+7=-83\)

th2 : \(x>7\) thì \(-37-\left|7-x\right|=-127\)

\(\Leftrightarrow\) \(-37-\left(x-7\right)=-127\) \(\Leftrightarrow\) \(-37-x+7=-127\)

\(\Leftrightarrow\) \(-x=-127+37-7=-97\Leftrightarrow x=97\)

vậy \(x=-83;x=97\)