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a) (x - 140) : 7 = 33 - 23 . 3
(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3
x - 140 = 3 x 7 = 21
x = 21 + 140 = 161
b) x3 . x2 = 28 : 23
x5 = 25
=> x = 2
c) (x + 2) . ( x - 4) = 0
x = -2 hoặc 4
d) 3x-3 - 32 = 2 . 32 =
3x-3 - 9 = 2 . 9 = 18
3x-3 = 18 + 9 = 27
3x-3 = 33
=> x - 3 = 3
x = 3 + 3 = 6
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Câu 2:
a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)
hay \(x\in\left\{1;7\right\}\)
\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)
\(\frac{62}{7}:x=\frac{1624}{27}\)
\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)
\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)
\(\frac{1}{5}:x=\frac{2}{35}\)
\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)
\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)
\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)
\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)
\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)
\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)
\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)
\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)
\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)
\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)
\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)
\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)
\(1-x=\frac{2}{5}:\frac{2}{5}=1\)
\(x=1-1=0\)
\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)
\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)
\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)
học tốt nha
a: x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b: 2x(x+3)=0
=>x(x+3)=0
=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
c: \(\left(6-x\right)\left(x+10\right)=0\)
=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)
d: \(\left(5x+20\right)\left(x^2+1\right)=0\)
=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)
=>5x=-20
=>x=-4
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a)\(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\times5}{5}=2\)
Vậy .............
b) \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{8\times6}{3}=16\)
Vậy ................
c) \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\times27}{9}=3\)
Vậy ................
d) \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\times6}{8}=3\)
Vậy ............
e) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+4x=20-6\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
f) \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy ...............
\(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\times5}{5}=2\)
\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{1\times27}{9}=3\)
\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{8\times6}{8}=6\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow3x+4x=20-6\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
1)(x-1)^2*(x-2)^2=0
=>(x-1)^2=0 hoặc (x-2)^2=0
=>x=1 hoặc x =2
Vậy x=1;x=2.
2)x(x+1)(x+2)^2(x+3)^3=0
=> x=0 hoặc x +1=0 hoặc x +2=0 hoặc x+3=0
=> x=0 ;x=-1;x=-2; x=-3
3)(x-9)^5(x+5)^8=0
=>x-9=0 hoặcx +5=0
=>x=9 hoặc x =0
4)(3x-9)^59(5x+75)^86=0
=>3x-9=0 hoặc 5x+75 =0
=>x=3 hoặc x= 15
Mình làm tắt một tí vì nó dựa vào cách làm câu 1 ấy !