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a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
a. 2x(x-5)-x(3+2x)=26
2\(x^2\)-10x-3x-2\(x^2\)=26
-13x=26
x=-2
b. (x-1)(2x-1)-(x-8)(x-1)=0
=> (x-1)(2x-1-x+8)=0
(x-1)(x+7)=0
x-1=0 hoặc x+7=0
x=1 hoặc x=-7
c)4x^2+4x+1=x^2-2x+1
4x^2-x^2+4x+2x=-1+1
3x^2+6x=0
3x(x+2)=0
3x=0 hoặc x+2=0
x=0 và x=-2