Các bạn yêu ơi giúp mik bài này vs
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\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
Con mèo vằn vào tranh, to hơn cả con hổ...dễ mến.(ko chắc nha, nếu đúng thì đây là so sánh hơn kém)
Bài 1:
a) \(3\dfrac{4}{5}.\left(-1\dfrac{1}{2}\right)+3\dfrac{4}{5}.2\dfrac{1}{2}-1\dfrac{4}{7}\)
\(=\dfrac{19}{5}.\left(\dfrac{-3}{2}+\dfrac{5}{2}\right)-\dfrac{11}{7}\)
\(=\dfrac{19}{5}.1-\dfrac{11}{7}\)
\(=\dfrac{19}{5}-\dfrac{11}{7}\)
\(=\dfrac{78}{35}\)
b) \(\dfrac{3}{4}:\left(-2\dfrac{1}{2}\right)-\dfrac{4}{5}:\left(-2\dfrac{1}{2}\right)+1\dfrac{1}{4}:\left(-2\dfrac{1}{2}\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{4}\right):\dfrac{-5}{2}\)
\(=\dfrac{6}{5}:\dfrac{-5}{2}\)
\(=\dfrac{-12}{25}\)
c) \(\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-1\dfrac{1}{3}\right)+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-\dfrac{4}{3}\right)+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.-2+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left[\dfrac{-6}{5}+\dfrac{4}{7}\right]\)
\(=\dfrac{4}{5}-\left(\dfrac{-22}{35}\right)\)
\(=\dfrac{10}{7}\)
Bài 2:
a) \(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}-\dfrac{4}{5}=1\dfrac{1}{5}\)
\(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=\dfrac{6}{5}+\dfrac{4}{5}\)
\(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=2\)
\(\dfrac{3}{4}-x=2:\dfrac{1}{2}\)
\(\dfrac{3}{4}-x=4\)
\(x=\dfrac{3}{4}-4\)
\(x=\dfrac{-13}{4}\)
b) \(\dfrac{3}{4}-\dfrac{1}{4}.x=2\dfrac{1}{3}\)
\(\dfrac{1}{4}.x=\dfrac{3}{4}-\dfrac{7}{3}\)
\(\dfrac{1}{4}.x=\dfrac{-19}{12}\)
\(x=\dfrac{-19}{12}:\dfrac{1}{4}\)
\(x=\dfrac{-19}{3}\)
Theo ĐLBT KL, có: mNa2SO4 + mBaCl2 = mNaCl + mBaSO4
⇒ mBaCl2 = mNaCl + mBaSO4 - mNa2SO4 = 11,7 + 23,3 - 14,2 = 20,8 (g)
Bạn tham khảo nhé!
a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{-x^2\left(x-2\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)