Tìm x\(\frac{x-1}{9}=\frac{8}{3}\)
b) \(\frac{x}{4}=\frac{18}{x+1}\)
c) \(\frac{-x}{4}=\frac{-9}{x}\)
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a/ \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-1\right)=72\)
\(\Leftrightarrow x-1=24\)
\(\Leftrightarrow x=25\)
Vậy ..
b/ \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x^2=6^2=\left(-6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy ..
c/ \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x\left(x+1\right)=72\)
\(\Leftrightarrow x\left(x+1\right)=8.9\)
\(\Leftrightarrow x=8\)
Vậy ..
a, \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=72:3\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=24+1\)
\(\Rightarrow x=25\)
b, \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow-x.x=-9.4\)
\(\Rightarrow-\left(x^2\right)=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
c, \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.18\)
\(\Rightarrow x.x+x.1=72\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+x-8^2+8=0\)
\(\Rightarrow x=8\)
a)\(\frac{x-1}{9}=\frac{8}{3}\)
(x-1)3=8.9
3x-3=72
3x=72+3
3x=75
x=75:3
x=25
b)\(\frac{-x}{4}=\frac{-9}{x}\)
(\(\frac{-x}{4}=\frac{-9}{x}\))=(\(\frac{x}{4}=\frac{9}{x}\))
x.x=9.4
x^2=36
x^2=6^2
➤x=6
c)\(\frac{x}{4}=\frac{18}{x+1}\)
x(x+1)=18.4
x(x+1)=72
(Ta có x và x+1 là hai số tự nhiên liên tiếp và một chẵn và một số lẻ)
Ta có:Ư(72)=1;2;3;4;6;8;9;72;36;12;18;24
Và vì x và x+1 là 2 số tự nhiên liên tiếp, nên:
Ta có bảng sau
x
|
1 | 3 | 8 | ||
x+1 | 2 | 4 | 9 | ||
x(x+1) | 2 | 12 | 72 | ||
d)\(\frac{5}{12}=\frac{-x}{72}\)
(-x).12=5.72
(-x).12=360
(-x)=360:12
(-x)=30
➤x=-30
e)\(\frac{x+3}{-15}=\frac{1}{3}\)
(x+3)3=1.(-15)
3x+9=-15
3x=(-15)-9
3x=-24
x=(-24):3
➤x=-8
mik ko chép lại đề, mik làm luôn:
a) x - \(\frac{31}{36}=\frac{-13}{38}\)
x = \(\frac{-13}{18}+\frac{31}{36}\)
\(x=\frac{5}{36}\)
b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)
\(\frac{11}{7}-x=\frac{3}{7}\)
x = \(\frac{11}{7}-\frac{3}{7}\)
x = 8/7
c) x + 3/11 = 23/44
x = 23/44 - 3/11
x = 1/4
d) \(\frac{1}{12}-x=\frac{-11}{9}\)
x = \(\frac{1}{12}+\frac{11}{9}\)
x = 47/36
e) \(x-\frac{2}{3}=\frac{-17}{3}\)
x= -17/3 + 2/3
x = -5
f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)
x - 1/2 = 3/4
x = 3/4 + 1/2
x = 5/4
g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)
2x + 3/8 = -3 / 8
2x = -3/8 - 3/8
2x = -9/8
x = -9/8.1/2
x = -9/16
h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)
x - \(\frac{x}{3}=\frac{1}{12}\)
x = \(\frac{1}{12}+\frac{x}{3}\)
x = \(\frac{1+4x}{12}\)
=> 12x = 1+4x
12x - 4x = 1
8x = 1
x = 1/8
e) \(\frac{-84}{14}< 3x< \frac{108}{9}\)
\(\Rightarrow-2< x< 4\)
Vậy \(-2< x< 4\)
d) \(\frac{x-3}{y-3}=\frac{4}{3}\)
Ta có: x=y+5
\(\Rightarrow\frac{y+5-3}{y-3}=\frac{4}{3}\)
\(\Rightarrow\frac{y+2}{y-3}=\frac{4}{3}\)
\(\Rightarrow\left(y+2\right).3=\left(y-3\right).4\)
\(3y+6=4y-12\)
\(6+12=4y-3y\)
\(y=18\)
\(\Rightarrow x=5+y=5+18=23\)
KL:............................
a) \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\)\(\left(x-1\right).3=9.8\)
\(3x-3=72\)
\(3x=75\)
\(x=\frac{75}{3}\)
\(x=25\)
Câu b tương tự
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x.\left(x+1\right)=4.18\)
\(x.\left(x+1\right)=72\)
\(x.\left(x+1\right)=8.9\)
\(\Rightarrow x=8\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\left(x-1\right)\cdot3=8\cdot9\)
\(\left(x-1\right)\cdot3=72\)
\(x-1=\frac{72}{3}\)
\(x-1=24\)
\(x=24+1\)
\(x=25\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(x\cdot\left(x+1\right)=18\cdot4\)
\(x\left(x+1\right)=72\)
\(x\left(x+1\right)=8\cdot9\)
\(x=8\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\frac{x}{-4}=\frac{-9}{x}\)
\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)
\(x^2=36\)
\(x^2=\left(-6\right)^2\)hoặc \(x^2=6^2\)
\(x=-6\) hoặc\(x=6\)
a)x-1/9=24/9 => x-1=24 =>x=23
b)x(x+1)=18*4 =>x=8
c)-x:4=-9:x =>-1.x2=-1.36 =>x=6
k mik nha!