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Bài 1:
a) \(2x+3=5\)
\(\Leftrightarrow2x=5-3=2\)
\(\Leftrightarrow x=2:2=1\)
Vậy: x=1
b) \(\left(2x-3\right)^2=9\)\(=3^2=\left(-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(3+3\right):2=3\\x=\left(-3+3\right):2=0\end{matrix}\right.\)
Vậy: \(x\in\left\{3;0\right\}\)
c) \(3x+4=2x-7\)
\(\Leftrightarrow2x-3x=7+4\)
\(\Leftrightarrow-1x=11\)
\(\Leftrightarrow x=-11\)
Vậy: x=-11
d) \(\left(4x-1\right)^3=27\)\(=3^3\)
\(\Rightarrow4x-1=3\)
\(\Leftrightarrow4x=3+1=4\)
\(\Leftrightarrow x=4:4=1\)
Vậy: x=1
e) \(\left(x-7\right).2=3.\left(2x+4\right)\)
\(\Leftrightarrow2x-14=6x+12\)
\(\Leftrightarrow2x-6x=12+14\)
\(\Leftrightarrow-4x=26\)
\(\Leftrightarrow x=\frac{26}{-4}=\frac{13}{-2}\)
Vậy: \(x=\frac{13}{-2}\)
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 !
\(a,x:\left(\frac{-1}{3}\right)^2=\frac{1}{3}\)
\(x:\left(\frac{1}{3}\right)^2=\frac{1}{3}\)
\(x=\frac{1}{3}\times\left(\frac{1}{3}\right)^2=\frac{1}{27}\)
\(b,\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)
\(x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)
\(x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{-1}{4}\)
\(d,\left(3x+1\right)^3=-27\)
\(\left(3x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+1=-3\)
\(3x=-4\)
\(x=\frac{-4}{3}\)
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
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/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)