Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Do:\left|x-12\right|=\left|12-x\right|\)
⇒2014.|x−12|+(x-12)2=2013.\(\left|x-12\right|\)
⇒2014.\(\left|x-12\right|\)+(x-12)2-2013.\(\left|x-12\right|\)=0
⇒(2014-2013).\(\left|x-12\right|\)+(x-12)2=0
⇒\(\left|x-12\right|+\left(x-12\right)^2\)=0
Do: \(\left|x-12\right|\ge0,\left(x-12\right)^2\ge0\)
⇒x-12=0
⇒x=12
I , tìm x :
a, \(\left|x\right|=1,21\)
Ta có : \(\left|x\right|=\left|1,21\right|\rightarrow\left|x\right|=\pm1,21\)
b, \(\dfrac{11}{12}-\left(\dfrac{2}{5}-x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{1}{4}\) => \(x=\dfrac{2}{5}-\dfrac{1}{4}\)
=> \(x=\dfrac{3}{20}\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}\div x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}\div x=\dfrac{-7}{20}\) => \(x=\dfrac{1}{4}\div\dfrac{-7}{20}\)
=> \(x=\dfrac{-5}{7}\)
d,\(3^x=81\)
Ta có 81= \(3^4\)
Vì : \(3^x=3^4\Rightarrow x=4\)
e,\(\dfrac{1}{2}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\left|x\right|-\dfrac{5}{6}=\dfrac{8}{3}:\dfrac{1}{2}\)
=> \(\left|x\right|-\dfrac{5}{2}=\dfrac{16}{3}\) => \(\left|x\right|=\dfrac{16}{3}+\dfrac{5}{2}\)
=> \(\left|x\right|=\dfrac{47}{6}\)
Vì \(\left|x\right|=\left|\dfrac{47}{6}\right|\Rightarrow x=\pm\dfrac{47}{6}\)
f, \(2^{x-3}=4\)
\(2^{x-3}=2^2\)
=> \(x-3=2\)
=> \(x=5\)
a, Ta có \(\left|x\right|=1,21\)
\(\Rightarrow\left[{}\begin{matrix}x=1,21\\x=-1,21\end{matrix}\right.\)
Vậy \(x\in\left\{1,21;-1,21\right\}\)
a) Thiếu đề (hoặc sai)
b) x đâu?
c)\(3x-1=x+2\)
\(\Rightarrow3x-x=2+1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
c) \(\frac{x+2}{5}=\frac{2-3x}{3}\)
\(\Rightarrow3.\left(x+2\right)=5.\left(2-3x\right)\)
\(\Rightarrow3x+6=10-15x\)
\(\Rightarrow3x+15x=10-6\)
\(\Rightarrow18x=4\)
\(\Rightarrow x=\frac{4}{18}=\frac{2}{9}\)
câu 1 là \(x\times\left(4.6+\frac{3}{5}\right)=7.2-8.15\)
câu 2 là \(42+\frac{3}{7}.\left[3\times x-1=12\right]\)
\(1.\)
Ta có :
\(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(y+z=-x\)
\(x+z=-y\)
\(\Rightarrow M=\left(-z\right)\left(-x\right)\left(-y\right)=-xyz\)
Mà \(xyz=2\)
\(\Rightarrow M=-2\)
Vậy : \(M=-2\)
\(2.\)
\(a.\)
Ta có :
\(yt.yz=48.24\)
\(\Rightarrow y^2.zt=48.24\)
Mà \(yt=32\Rightarrow y^2.32=48.24\)
\(\Rightarrow y^2=\frac{48.24}{32}\)
\(\Rightarrow y^2=36\)
\(\Rightarrow y=\pm6\)
+ Nếu \(x=6\)
Ta có : \(t=48:6=8\)
\(z=24:6=4\)
\(x=12:6=2\)
+ Nếu \(y=-6\)
Ta có : \(t=48:\left(-6\right)=-8\)
\(z=24:\left(-6\right)=-4\)
\(x=12:\left(-6\right)=-2\)
Vậy \(x=-2;y=-6;z=-4;t=-8\) hoặc \(x=2;y=6;z=4;t=8\)
\(b.\)
Ta có :
\(y+t=11\) \(\left(1\right)\)
\(y+z=9\) \(\left(2\right)\)
\(x+y=6\) \(\left(3\right)\)
\(z+t=12\) \(\left(4\right)\)
Lấy \(\left(1\right)+\left(2\right)\), ta được :
\(2y+t+z=20\)
Mà \(t+z=12\)
\(\Rightarrow2y+12=20\)
\(\Rightarrow2y=8\)
\(\Rightarrow y=4\)
Từ \(\left(2\right)\) \(\Rightarrow z=9-y=9-4=5\)
Từ \(\left(3\right)\) \(\Rightarrow x=6-y=6-4=2\)
Từ \(\left(4\right)\) \(\Rightarrow t=12-z=12-5=7\)
Vậy : \(x=2;y=4;z=5;t=7\)
x = 12 nha bạn
vì sao