Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) \(\left(3\times x-15\right)^7=0.\)
\(3\times x-15=0\)
\(3\times x=15\)
\(x=5\)
b ) \(10-\left\{\left[\left(x\div3+17\right)\div10+3\times2^4\right]\div10\right\}=5\)
\(10-\left\{\left[\left(x\div3+17\right)\div10+3\times16\right]\div10\right\}=5\)
\(10-\left\{\left[\left(x\div3+17\right)\div10+48\right]\div10\right\}=5\)
\(\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\left(x\div3+17\right)\div10+48=50\)
\(\left(x\div3+17\right)\div10=2\)
\(x\div3+17=20\)
\(x\div3=3\)
\(x=9\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(3\left(2x-\frac{5}{4}\right)=\left(3-1\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)
\(\Leftrightarrow6x-\frac{15}{4}=\frac{3}{2}x+\frac{1}{12}\)
\(\Leftrightarrow\frac{9}{2}x+\frac{3}{4}=\frac{15}{4}\)
\(\Leftrightarrow\frac{9}{2}x=3\)
\(\Leftrightarrow x=\frac{2}{3}\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
Bài 2
\(a,\)\(\left(x^2+7\right)\left(x^2-49\right)< 0\)
Vì \(x^2+7>0\)\(\Rightarrow x^2-49< 0\)
\(\Rightarrow\left(x-7\right)\left(x+7\right)< 0\)
\(...\)
Bài 2:
a) \(\left(x^2+7\right).\left(x^2-49\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+7< 0\\x^2-49>0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+7>0\\x^2-49< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2< -7\\x^2>49\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x^2>-7\\x^2< 49\end{cases}}\)
\(\Leftrightarrow-7< x^2< 49\)
Mà \(x^2\ge0\)và \(x^2\)là 1 SCP
\(\Rightarrow x^2\in\left\{1;4;9;16;25;36\right\}\)
\(\Rightarrow x\in\left\{1;2;3;4;5;6\right\}\)
Vậy \(x\in\left\{1;2;3;4;5;6\right\}\)
6 va 65