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1) -3-(-x) =5
=> -3 + x = 5
=> x = 5 - (-3)
=> x = 8
2)-9-(x-3)=12
=> x- 3 = (-9) - 12
=> x-3 =-21
=> x = (-21)+3
=> x = -18
3)-10-(x-5)=-5
x-5 = (-10) -(-5)
x-5 = -5
x = -5 +5
x = 0
4)210-(10-x) =110
10-x = 210 -110
10-x = 100
x = 10 -100
x = -90
5)4x-(2x-5)=21
=> 4x-2x + 5 =21
=> x(4-2) + 5 = 21
=> x.2 =21 - 5
=> x.2 = 16
=> x = 16:2
=> x = 8
6)14x-5=8x+10
7)15+5x =3x+30
8)2x-5=15-3x
9)2(3x+5)+3(x+1)=x+220
10)4.(x-2+5(x-3)=18
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận thấy \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
=> \(\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> \(2x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{6}\)
Vậy Min A = -1 <=> X = -1/6
a, \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\)
Dấu "=" xảy ra <=> 2x+1/3=0 <=> x= -1/6
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\(12-(23-17)+\left|-15\right|=13-x-(23-8+\left|-9\right|)\)
\(\Rightarrow12-6+15=13-x-(23-8+9)\)
\(\Rightarrow12-6+15=13-x-24\)
\(\Rightarrow21=13-x-24\)
\(\Rightarrow x=13-24-21\)
\(\Rightarrow x=-32\)
\(\text{Vậy : }x=-32\)
12-( 23 - 17 )+ 15 = 13 -x - (23 - 8 + 9 )
12-23+17+15=13-x-23+8-9
12+17+15-23=13+8-9-x
29+15-23= 21-9 -x
44-23=12-x
21=12-x
12-x=21
x=12-21
x= -9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=\dfrac{86}{2}\\ x=43\)
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=15^{10}:3^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=5^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=86:2\\ x=43\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x+\left(-21\right)=12-8x\)
\(3x-21=12-8x\)
\(3x+8x=12+21\)
\(11x=33\)
\(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6\cdot x+9=15\)
\(\Rightarrow6\cdot x=15-9\)
\(\Rightarrow6\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{6}=1\)
_______________
\(43+2x=49\)
\(\Rightarrow2x=49-43\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
____________
\(x:2+5=11\)
\(\Rightarrow x:2=11-5\)
\(\Rightarrow x:2=6\)
\(\Rightarrow x=6\cdot2=12\)
______________
\(77-11x=0\)
\(\Rightarrow11x=77\)
\(\Rightarrow x=\dfrac{77}{11}\)
\(\Rightarrow x=7\)
_______________
\(12-4:x=8\)
\(\Rightarrow4:x=12-8\)
\(\Rightarrow4:x=4\)
\(\Rightarrow x=\dfrac{4}{4}=1\)
_____________
\(x:3+8=11\)
\(\Rightarrow x:3=11-8\)
\(\Rightarrow x:3=3\)
\(\Rightarrow x=3\cdot3=9\)
a: 6x+9=15
=>6x=6
=>x=1
b: 2x+43=49
=>2x=6
=>x=3
c: x:2+5=11
=>x:2=6
=>x=12
d: 77-11x=0
=>7-x=0
=>x=7
e: 12-4:x=8
=>4:x=4
=>x=1
f: x:3+8=11
=>x:3=3
=>x=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
a, Có thể lập xy=21 <=> x=3;y=7 hoặc x=-3;y=-7
<=> x=7;y=3 hoặc x=-7;y=-3 ....v..v...
b, \(\left(x+5\right)\left(y-3\right)=15\)
\(\Rightarrow\orbr{\begin{cases}x+5=15\\y-3=15\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\y=18\end{cases}}}\)
c, \(\left(2x-1\right)\left(y-3\right)=12\)
\(\Rightarrow\orbr{\begin{cases}2x-1=12\\y-3=12\end{cases}\Rightarrow\orbr{\begin{cases}2x=13\\y=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{13}{2}\\y=15\end{cases}}}\)
Bài 2
Ư(6)={1;2;3;6} => 1+2+3+6=12
Ư(8)={1;2;4;8} => 1+2+4+8 =15
=> Tổng 2 ước này đều \(⋮3\)
๖²⁴ʱミ★Šїℓεŋէ❄Bʉℓℓ★彡⁀ᶦᵈᵒᶫ mù mắt =)) t làm mẫu câu b thôi, c nhìn vào mà làm
b) \(\left(x+5\right)\left(y-3\right)=15\)
\(\Rightarrow y-3=\frac{15}{x+5}\Rightarrow y=3+\frac{15}{x+5}\)
\(\Rightarrow x+5\inƯ\left(15\right)\)
Ta có: \(Ư\left(15\right)=\left\{-15;-5;-3;-1;0;1;3;5;15\right\}\)
\(x=\left\{0;-10;-8;-6;-20;-4;-2;0;10\right\}\)
Vì \(x\inℕ\Rightarrow x=\left\{0;10\right\}\)
\(\Rightarrow y=\left\{6;4\right\}\)
Vậy: (x,y) = {(0;10); (6;4)}
2x + 8x = (-15) + (-9)
<=> 10x = -24
<=> x = -2,4
2x +8x=(-15)+(-9)
2x +8x=-24
x(2+8)=-24
x. 10=-24
x=-24:10
x=-2,4