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a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)

a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x+3^x.3^2=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x.8-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x.7=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)

a) Đề sai.
b) \(\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0-1\\\sqrt{x}=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=9\end{matrix}\right.\)
Vậy \(x=9.\)
c) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
Chúc bạn học tốt!

\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\Rightarrow x-\frac{1}{2}=\frac{1}{3}\Rightarrow x=\frac{5}{6}\)
(x-1/2)^3=1/27
(X-1/2)^3=(1/3)^3
X-1/2=1/3
X =1/3+1/2
X =2/6+3/6
X =5/6

Lời giải :
Theo đề bài ta có \(\frac{x}{\frac{5}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{6}{5}}\Leftrightarrow\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}\)
Đặt \(\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{5k}{2}\\z=\frac{6k}{5}\end{cases}}\)
Mặt khác : \(\frac{x}{2}=\frac{z-28}{3}\)
\(\Leftrightarrow3x-2z=-56\)
\(\Leftrightarrow3\cdot\frac{5k}{2}-2\cdot\frac{6k}{5}=-56\)
\(\Leftrightarrow k=\frac{-560}{51}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1400}{51}\\y=\frac{-2240}{153}\\z=\frac{-224}{17}\end{cases}}\)
\(B=x+y-z=\frac{-1400}{51}+\frac{-2240}{153}-\frac{-224}{17}=\frac{-4424}{153}\)

a. Thay \(x=-\frac{2}{3}\) vào \(C=6x^3-3x^2+2\left|x\right|+4\), ta có :
\(C=6\left(-\frac{2}{3}\right)^3-3\left(-\frac{2}{3}\right)^2+2\left|-\frac{2}{3}\right|+4\)
\(\Rightarrow C=6.\frac{-8}{27}-3.\frac{4}{9}+2.\frac{2}{3}+4\)
\(\Rightarrow C=-\frac{16}{9}-\frac{4}{3}+\frac{8}{3}+4\)
\(\Rightarrow C=\frac{32}{9}\)
b. Thay \(x=\frac{1}{2};y=-3\)vào \(D=2\left|x\right|-3\left|y\right|\), ta có :
\(D=2\left|\frac{1}{2}\right|-3\left|-3\right|\)
\(\Rightarrow D=2.\frac{1}{2}-3.3\)
\(\Rightarrow D=2-9\)
\(\Rightarrow D=-7\)

x2y + 0,5xy3 - 7,5x3y2 + x3 + 3xyz2 - x2y + 5,5x3y2
= 0,5xy3 -2x3y2 + x3+ 3xyz2
Ta co : \(3^x+3^{x+2}=2430\)
\(3^x.1+3^x.3^2=2430\)
\(3^x\left(1+3^2\right)=2430\)
\(3^x.10=2430\)
\(3^x=2430:10\)
\(3^x=243\)
\(\Rightarrow3^x=3^5\)
Vay x=5
**** nhe
\(3^x+3^{x+2}=2430\)
=> \(3^x.\left(1+3^2\right)=2430\)
=>\(3^x.10=2430\)
=>\(3^x=243\)
=>\(3^x=3^5\)
Vậy \(x=5\).