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\(a,\Rightarrow6x+70=130\Rightarrow6x=60\Rightarrow x=10\\ b,\Rightarrow240=\left(x+70\right):14-40\\ \Rightarrow\left(x+70\right):14=280\\ \Rightarrow x+70=3920\Rightarrow x=3850\\ c,\Rightarrow x-15=75\Rightarrow x=90\\ d,\Rightarrow\left(x+175\right):5=680-30=650\\ \Rightarrow x+175=3250\Rightarrow x=3075\\ e,\Rightarrow x-4867=1004523\Rightarrow x=1009390\\ f,\Rightarrow x+32-17=24\Rightarrow x=9\\ g,\Rightarrow3\left(x+1\right)=54\Rightarrow x+1=18\Rightarrow x=17\\ h,\Rightarrow19\left(35:x+3\right)=152\\ \Rightarrow35:x+3=8\Rightarrow35:x=11\Rightarrow x=\dfrac{35}{11}\)
\(a,\dfrac{5}{8}=\dfrac{x}{14}\)
\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)
Vậy \(x=8,75\)
\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)
Vậy \(x=-2\)
\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)
\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)
Vậy \(x=-6\)
câu d đã có đáp án
a) 2436 : x = 12
\(\Rightarrow\) x = 2436 : 12
\(\Rightarrow\) x = 203
b) 6x - 5 = 613
\(\Rightarrow\) 6x = 618
\(\Rightarrow\) x = 103
c) 12(x - 1) = 0
\(\Rightarrow\) x - 1 = 0
\(\Rightarrow\) x = 1
d) 0 : x = 0 (đúng \(\forall\)x \(\in\) N)
\(\Rightarrow\) x \(\in\) N
a) Ta có: 2436:x=12
nên x=2436:12
hay x=203
b) Ta có: 6x-5=613
nên 6x=618
hay x=103
c) Ta có: 12(x-1)=0
mà 12>0
nên x-1=0
hay x=1
d) Ta có: 0:x=0
nên \(x\in R;x\ne0\)
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
a,x.(3\4+2\5)=1
x.20\23=1
x=1:20\23
x=20\23
b,x-9\11=0 hoặc x-25\31=0
x=9\11 x=25\31
c,x-3\7.9\14=7\3
x-2\3=7\3
x=7\3+2\3
x=9\3
x=3
a) x+(x+1)+(x+2)+...+(x+2010)=2029099
<=>2011x+(1+2+3+...+2010)=2029099
<=>2011x+2021055=2029099
<=>2011x = 8044 <=> =4
b) 5x-14=x+34
<=> 5x-x = 34+14
<=> 4x = 48 => x = 12
c) (x+5) .(x+9) =0 <=>
x+5 = 0 --> x = -5
hoặc x+9 = 0 --> x = -9
a/ x=4
b/ x=12
c/ x=-5 hoặc x=-9