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13) \(720\div\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Rightarrow720\div\left[41-\left(2x-5\right)\right]=40\)
\(\Rightarrow41-\left(2x-5\right)=720\div40\)
\(\Rightarrow41-\left(2x-5\right)=18\)
\(\Rightarrow2x-5=41-18\)
\(\Rightarrow2x-5=23\)
\(\Rightarrow2x=23+5\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=28\div2\)
\(\Rightarrow x=14\)
14) \(\left\{\left[\left(2x+14\right)\div2^2-3\right]\div2\right\}-1=0\)
\(\left[\left(2x+14\right)\div2^2-3\right]\div2=0+1\)
\(\left[\left(2x+14\right)\div2^2-3\right]\div2=1\)
\(\left(2x+14\right)\div2^2-3=1.2\)
\(\left(2x+14\right)\div2^2-3=2\)
\(\left(2x+14\right)\div2^2=2+3\)
\(\left(2x+14\right)\div2^2=5\)
\(2x+14=5.2^2\)
\(2x+14=20\)
\(2x=20-14\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)

a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(1999^{2x-6}=1\)
\(\Rightarrow1999^{2x-1}=1999^0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
c) \(x^{2002}=x\)
\(\Rightarrow x^{2002}-x=0\)
\(\Rightarrow x.\left(x^{2001}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)
+) \(x=0\)
+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d) \(\left(x-1\right)^2=9\)
\(\Rightarrow x-1=\pm3\)
+) \(x-1=3\Rightarrow x=4\)
+) \(x-1=-3\Rightarrow x=-2\)
Vậy \(x\in\left\{4;-2\right\}\)
e) \(\left(2x-3\right)^2=81\)
\(\Rightarrow2x-3=\pm9\)
+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)
+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)
Vậy \(x\in\left\{6;-3\right\}\)
Các phần khác làm tương tự

Giải:
a) \(\left(4,5-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\left(-\dfrac{3}{7}\right)=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\left(-\dfrac{3}{7}\right)=-\dfrac{11}{6}\)
\(\Leftrightarrow2x=4,5-\left(-\dfrac{11}{6}\right)\)
\(\Leftrightarrow2x=\dfrac{19}{3}\)
\(\Leftrightarrow x=\dfrac{19}{3}:2=\dfrac{19}{6}\)
Vậy ...
b) \(\dfrac{4}{9}x=\dfrac{9}{8}-0,125\)
\(\Leftrightarrow\dfrac{4}{9}x=\dfrac{9}{8}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{9}x=1\)
\(\Leftrightarrow x=1:\dfrac{4}{9}=\dfrac{9}{4}\)
Vậy ...
Các câu còn lại làm tương tự.

a) x \(\vdots\) 6 => x \(\in\) Ư(6) = {1;2;3;6}
Mà 6 \(\vdots\) x - 1 => x \(\in\) {2;3;4;7}
Câu còn lại bn tự lm nha
a)
6 : x - 1
=> x - 1 thuộc Ư(6) = { 1; 2; 3; 6; -1; -2; -3; -6 }
Sau đó lập bảng tìm x
b)
tương tự
^^

a, \(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy......
b, \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy......
c, \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
\(\Rightarrow\dfrac{1}{7}:x=\dfrac{-3}{14}\)
\(\Rightarrow x=\dfrac{1}{7}:\dfrac{-3}{14}=\dfrac{-2}{3}\)
Vậy.....
Chúc bạn học tốt!!!
a,(2x-3)(\(\dfrac{3}{4}\)x+1)=0
*2x-3=0\(\rightarrow\)2x=3\(\rightarrow\)x=\(\dfrac{3}{2}\)
*\(\dfrac{3}{4}\)x+1=0\(\rightarrow\)\(\dfrac{3}{4}\)x=\(-1\)\(\rightarrow\)x=\(\dfrac{-3}{4}\)
Vậy x\(\in\){\(\dfrac{-3}{4};\dfrac{3}{2}\)}
b,(5x-1)(2x-1/3)=0
*5x-1=0\(\rightarrow\)5x=1\(\rightarrow\)x=1/5
*2x-1/3=0\(\rightarrow\)2x=1/3\(\rightarrow\)x=1/6
Vậy x\(\in\){1/5;1/6}
c,3/7+1/7:x=3/14
1/7:x=-3/14

\(a.6⋮\left(x-1\right)\)
\(\Rightarrow x-1\inƯ\left(6\right)\)
\(\LeftrightarrowƯ\left(6\right)=\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{2;3;4;7\right\}\)
\(b.14⋮\left(2x+3\right)\)
\(\Rightarrow2x+3\inƯ\left(14\right)\)
\(\LeftrightarrowƯ\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow x\in\left\{2\right\}\)

14⋮ (2x+3)
=> (2x+3) ∈ Ư(14)= { -14;-7;-2;-1;1;2;7;14}
ta có bảng sau
2x+3 | -14 | -7 | -2 | -1 | 1 | 2 | 7 | 14 |
2x | -17 | -10 | -5 | -4 | -2 | -1 | -4 | -11 |
x | -17/2 | -5 | -5/2 | -2 | -1 | -1/2 | -2 | -11/2 |
vậy .....

a, \(2-\dfrac{14}{x}=\dfrac{-22}{3}\)
\(\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\dfrac{14}{x}=\dfrac{28}{3}\)
=> \(x.28=14.3\)
\(x.28=42\)
\(x=42:28\)
\(x=\dfrac{3}{2}=1,5\)
b, \(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)=-\dfrac{1}{5}\)
\(\dfrac{2x}{5}=-\dfrac{1}{5}-1=\dfrac{-6}{5}\)
\(\dfrac{2x}{5}=\dfrac{-6}{5}\)
=> \(2x=-6\)
\(x=-6:2=-3\)
a)
\(2-\dfrac{14}{x}=-\dfrac{22}{3}\)
\(\Rightarrow\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\Rightarrow x=\dfrac{14.3}{28}=\dfrac{3}{2}=1,5\)
b)
\(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\Rightarrow\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)\)
\(\Rightarrow\dfrac{2x}{5}+1=-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{2x}{5}=-\dfrac{1}{5}-1=-\dfrac{6}{5}\)
Hay \(\dfrac{2x}{5}=-\dfrac{6}{5}\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-\dfrac{6}{2}=-3\)
Chúc bạn học tốt!