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a,(x+1)-(x+2)-(x+3)=24
=>x+1-x-2-x-3 =24
=>(x-x-x)+(1-2-3) =24
=> -x-4 =24
=> -x =24+4
=> -x =28
=> x =-28
Vậy x=-28
b,4x+2-3(x-1)=3x-5
=>4x+2-3x+3=3x-5
=>3x-4x+3x =2+3+5
=>2x =10
=>x =5
Vậy x=5
c,x-1-2(x-2)=x-11
=>x-1-2x+4=x-11
=>x-2x-x =-11+1-4
=>-2x =-14
=>x =7
Vậy x = 7
=> 5 - [ 4 - ( 1 + 2x ) ] = -6
=> 4 - 1 - 2x = 11
=> 2x = 3 - 11 = -8
=> x = -4
X^3-X^2=X^2[X-1]=0
X^2=0 thì X=0
X-1=0 THÌ X=1
Vậy thỏa mãn được yêu cầu đề bài
\(\left(x-3\right)\cdot\left(y-5\right)=3\)
=>\(\left(x-3\right)\cdot\left(y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)
=>\(\left(x-3;y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(4;8\right);\left(6;6\right);\left(2;2\right);\left(0;4\right)\right\}\)
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
a) x+3=12
x=12-3
x=9
b)(x-3):2=514:512
=>(x-3):2=52
=>(x-3):2=25
=>x-3=25.2
=>x-3=50
=>x=50+3
=>x=53
c)4x+3x=30-20:10
=>x(4+3)=30-2
=>7x=28
=>x=28:7
=>x=4
d)2x-138=23.32
=>2x-138=8.9
=>2x-138=72
=>2x=72+138
=>2x=210
=>x=210:2
=>x=105
a) x + 3 = 12
x = 12 - 3
x = 9
b) ( x - 3 ) : 2 = 514 : 512
( x - 3 ) : 2 = 514-12
( x - 3 ) : 2 = 52
( x - 3 ) : 2 = 25
x - 3 = 50
x = 53
c) 4x + 3x = 30 - 20 : 10
7x = 28
x = 4
d) 2x - 138 = 23 x 32
2x - 138 = 8 x 9
2x - 138 = 72
2x = 210
x = 105
[(10-x).2+5]:3=3+2
\(\Rightarrow\)[(10-x).2+5]:3=5
(10-x).2+5=5.3
(10-x).2+5=15
(10-x).2=15-5
(10-x).2=10
10-x=10:2
10-x=5
\(\Rightarrow\)x=10-5
x=5
x = 5 nho k cho minh nhe