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1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
\(2x+3=8\)
\(\Rightarrow2x=8-3\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(x:5-2=3\)
\(\Rightarrow x:5=3+2\)
\(\Rightarrow x:5=5\)
\(\Rightarrow x=5\cdot5\)
\(\Rightarrow x=25\)
\(x:7-2=19\)
\(\Rightarrow x:7=19+2\)
\(\Rightarrow x:7=21\)
\(\Rightarrow x=21\cdot7\)
\(\Rightarrow x=147\)
Mình chưa rõ đề
\(20-\left(x+3\right)=5\)
\(\Rightarrow-x-3=5-20\)
\(\Rightarrow-x-3=-15\)
\(\Rightarrow-x=-15+3\)
\(\Rightarrow-x=-12\)
\(\Rightarrow x=12\)
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 41 - 31
5x = 10
x = 10 : 5
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 38 - 24
7x = 14
x = 14 : 7
x = 2
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 15 - 15
9x = 0
x = 0
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 22 - 20
x = 2
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 49 - 41
x = 8
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x - 11x + 8 = -10
-9x = -10 - 8
-9x = -18
x = -18 : (-9)
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 12 - 2
7x = 10
x = 10/7
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
10x - 15 = x + 22
10x - x = 22 + 15
9x = 37
x = 37/9
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
11x - 6 = 2x + 1
11x - 2x = 1 + 6
9x = 7
x = 7/9
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
6 + 4x = 35
4x = 35 - 6
4x = 29
x = 29/4
a. x:(1/2+2/3)=6/5
=>x:7/6=6/5
=>x=6/5*7/6=>x=7/5
b.(x-1/2)-5(x-2/3)=3/2x
=>x-1/2-5x+10/3=3/2x
=>-4x+17/6=3/2x
=>17/6x=3/2x--4x
=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33
c.-5(x+1/5)-1/2(x-2/3)=x
=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x
=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x
=>x=-4/39
a) x : (1/2 + 2/3) = 6/5
=> x : 7/6 = 6/5
=> x = 6/5 x 7/6
x = 7/5
b) , c) ko bít hihi
bài 2: (x-3).(y+2) = -5
Vì x, y \(\in\)Z => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}
Ta có bảng:
x-3 | 5 | -5 | -1 | 1 |
y+2 | 1 | -1 | -5 | 5 |
x | 8 | -2 | 2 | 4 |
y | -1 | -3 | -7 | 3 |
bài 3: a(a+2)<0
TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)
TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
Vậy -2<a<0
Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)
TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2
TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại
Vậy 1<a<2
Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.
\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)
Vậy \(\left(x;y\right)=\left(5;-2\right)\)