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\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)
\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)
\(a)\frac{5}{8}-x=2\frac{1}{6}\)
\(\Rightarrow\frac{5}{8}-x=\frac{13}{6}\)
\(\Rightarrow x=\frac{5}{8}-\frac{13}{6}\)
\(\Rightarrow x=\frac{15}{24}-\frac{52}{24}\)
\(\Rightarrow x=-\frac{37}{24}\)
\(b)\) \(\frac{4}{9}:x=-\frac{1}{3}+1\frac{1}{6}\)
\(\Rightarrow\frac{4}{9}:x=-\frac{1}{3}+\frac{7}{6}\)
\(\Rightarrow\frac{4}{9}:x=-\frac{2}{6}+\frac{7}{6}\)
\(\Rightarrow\frac{4}{9}:x=\frac{5}{6}\)
\(\Rightarrow x=\frac{4}{9}:\frac{5}{6}\)
\(\Rightarrow x=\frac{4}{9}.\frac{6}{5}\)
\(\Rightarrow x=\frac{8}{15}\)
\(c)\left(3x-2\right)^3=-\frac{1}{27}\)
\(\Rightarrow3x-2=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+2\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{6}{3}\)
\(\Rightarrow3x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:3\)
\(\Rightarrow x=\frac{5}{9}\)
d ) và e ) tự làm
Chúc bạn học tốt !!!
Câu 1:
a) Ta có: x-3 là ước của 13
\(\Leftrightarrow x-3\inƯ\left(13\right)\)
\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)
hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)
Vậy: \(x\in\left\{4;2;16;-10\right\}\)
b) Ta có: \(x^2-7\) là ước của \(x^2+2\)
\(\Leftrightarrow x^2+2⋮x^2-7\)
\(\Leftrightarrow x^2-7+9⋮x^2-7\)
mà \(x^2-7⋮x^2-7\)
nên \(9⋮x^2-7\)
\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)
\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)
mà \(x^2-7\ge-7\forall x\)
nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)
\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)
\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)
mà \(x\in Z\)
nên \(x\in\left\{2;-2;4;-4\right\}\)
Vậy: \(x\in\left\{2;-2;4;-4\right\}\)
Câu 2:
a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)
\(\Leftrightarrow2x-6-3x+15=12-4x-18\)
\(\Leftrightarrow-x+9+4x+6=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
a, 3x-(2x+1)=6
3x-2x-1=6
x-1=6
x=7
b,2x-[(-15)+x]-6=16
2x-[(-15)+x]=22
2x-(-15)-x=22
x+15=22
x=7
c,(15-18)2+3x=2.(x-6)
(-3)2+3x=2x-12
9+3x=2x-12
3x-2x=-12-9
x=-21
a: =>1/3x-2/5x=5
=>-1/15x=5
=>x=-75
b: =>4x=4
=>x=1
c: =>6*3^x-5*3^x=243
=>3^x=243
=>x=5
a: =>2x=-18+5=-13
=>x=-13/2
b: =>3^x-1=81
=>x-1=4
=>x=5
c: =>4(5-x)=24
=>5-x=6
=>x=-1
a) \(\left|18-2x\right|=18\Rightarrow\orbr{\begin{cases}18-2x=18\\18-2x=-18\end{cases}\Rightarrow\orbr{\begin{cases}2x=0\\2x=36\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=18\end{cases}}}\)
b) \(\left|1-x\right|=1-x\Rightarrow\orbr{\begin{cases}1-x=1-x\\1-x=-\left(1-x\right)=-1+x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left\{\mp1;\mp2;...\right\}\\-x-x=-1-1\Rightarrow x=1\end{cases}}\)
Làm từng nãy đã , mỏi tay
a) ta có \(|\)18 - 2x\(|\)=18
=> 18-2x = 18 hoặc 18 -2x=-18
-2x = 0 -2x = -36
x = 0 x = 18
vậy x = 0 hoặc x=18