a) x + 10 = 20

<=> x = 20 - 10 = 10

Vậy x = 10

b) 2x + 15 = 35

<=> 2x = 35 - 15 = 20

<=> x = 10

Vậy x = 10

c) 3(x + 2) = 15

<=> x + 2 = 15 : 3 = 5

<=> x = 5 - 2 = 3

Vậy x = 3

d) 10x + 15.11 = 20.10

<=> 10x + 165 = 200

<=> 10x = 200 - 165 = 35

<=> x = 35 : 10 = 3,5

Vậy x = 3,5

e) 4(x + 2) = 3.4

<=> x + 2 = 3

<=> x = 3 - 2 = 1

Vậy x = 1

f) 33x + 135 = 26.9

<=> 33x + 135 = 234

<=> 33x = 234 - 135 = 99

<=> x = 99 : 33 = 3

Vậy x = 3

g) 2x + 15 + 16 + 17 = 100

<=> 2x + 48 = 100

<=> 2x = 100 - 48 = 52

<=> x = 52 : 2 = 26

Vậy x = 26

h) 2(x + 9 + 10 + 11) = 4.12.5

<=> x + 30 = 120

<=> x = 120 - 30 = 90

Vậy x = 90

thiên tài cũng phát ngất

B2: tính:

a, 225-228:[4.(48-72)]

=225-228:[4.(-24)]

=225-228:(-96)

=225-(-19/8)

=1819/8

5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)

a) 

Ta có : vì|1/2-1/3+x| lớn hơn hoặc bằng 0 

Còn -1/4-|y| bé hơn hoặc bằng 0

=> ko tồn tại x

b) 

Ta có: |x-y| lớn hơn hoặc bằng 0 và|y+9/25| lớn hơn hoặc bằng 0 mà:

| x-y|+ |y+9/25| =0 => |x-y| =0 và |y+9/25|=0

 Xét |y+9/25| có:

| y+9/25|=0 => y+9/25=0 => y=-9/25

Thay y = -9/25 vào |x-y| =0 => x=-9/25

 Vậy x=y=-9/25

#maianhhappy

bài 1 tính giá trị biểu thức

( - 25 ) nhân ( -3 ) nhân x với x = 4

\(\left(-25\right).\left(-3\right).4\)

\(=\left(-25\right).4.\left(-3\right)\)

\(=-100.\left(-3\right)=300\)

( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25

\(\left(-1\right).\left(-4\right).5.8.25\)

\(=4.5.8.25=4.25.5.8\)

\(=100.40=40000\)

( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12

\(\left(2.4.\left(-6\right)\right)^2:12\)

\(=\left(-48\right)^2:12\)

\(=2304:12=192\)

[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9

\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)

\(=-2700:\left(-9\right)\)

\(=300\)

(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3

\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)

\(=16:2.8\)

\(=8.8=64\)

bài 2 tìm x

( 2x _ 5 ) + 17 = 6

\(2x-5=-11\)

\(2x=-6\)

\(x=-3\)

10 _ 2 ( 4 _ 3x ) = - 4

\(2.\left(4-3x\right)=14\)

\(4-3x=7\)

\(3x=-3\)

\(x=-1\)

- 12 + 3 ( -x + 7 ) = -18

\(3\left(-x+7\right)=-6\)

\(-x+7=-2\)

\(-x=-9\)

\(x=9\)

24 : ( 3x _ 2 ) = -3

\(3x-2=-8\)

\(3x=-6\)

\(x=-2\)

- 45 : 5 nhân ( -3 _ 2x ) = 3

\(5.\left(-3-2x\right)=-15\)

\(-3-2x=-3\)

\(2x=0\)

\(x=0\)

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

2: Tìm x

a) Ta có: x+25=40

nên x=40-25=15

Vậy: x=15

b) Ta có: 198-(x+4)=120

\(\Leftrightarrow x+4=198-120=78\)

hay x=78-4=74

Vậy: x=74

c) Ta có: \(\left(2x-7\right)\cdot3=125\)

\(\Leftrightarrow2x-7=\dfrac{125}{3}\)

\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)

\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)

Vậy: \(x=\dfrac{73}{3}\)

d) Ta có: \(x+16⋮x+1\)

\(\Leftrightarrow x+1+15⋮x+1\)

mà \(x+1⋮x+1\)

nên \(15⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(15\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)

Tự kết luận nhé bạn