Có nhớ chăng hỡi gió rét thành Ba Lê
Một viên gạch hồng, Bác chống lại cả mùa đong giá rét.

Những giọt sương sớm long lanh còn đang đọng lại trên lá trông như những hạt pha lê thủy tinh

\(y^2-6y+9=0\\\left(y-3\right)^2=0\\ y-3=0\\ y=3\)

=> C

a. \(-\dfrac{3}{4}x+\dfrac{7}{11}=\dfrac{3}{22}\\ -\dfrac{3}{4}x=\dfrac{3}{22}-\dfrac{7}{11}\\ -\dfrac{3}{4}x=-\dfrac{11}{22}\\ x=-\dfrac{11}{22}:-\dfrac{3}{4}\\ x=\dfrac{22}{33}\)

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b. \(\left|1\dfrac{1}{3}-0,4x\right|-\dfrac{5}{12}=\dfrac{1}{4}\\ \left|\dfrac{4}{3}-0,4x\right|=\dfrac{1}{4}+\dfrac{5}{12}\\ \left|\dfrac{4}{3}-0,4x\right|=\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{4}{3}-0,4x=\dfrac{2}{3}\\\dfrac{4}{3}-0,4x=-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0,4x=\dfrac{4}{3}-\dfrac{2}{3}\\-0,4x=\dfrac{4}{3}--\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}:0,4\\x=2:-0,4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-5\end{matrix}\right.\)

a. \(\left(x-3\right)\left(2x+1\right)=10\\ \Rightarrow\left[{}\begin{matrix}x-3=10\\2x+1=10\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=13\\x=\dfrac{9}{2}\end{matrix}\right.\)

b. \(\left(x-3\right)\left(y+1\right)=-8\\ \Rightarrow\left[{}\begin{matrix}x-3=8\\y+1=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=11\\y=7\end{matrix}\right.\)

\(\left(x+3\right)^2=16\\ \left(x+3\right)^2=4^2\\ x+3=4\\ x=1\)

\(\left(12x-5\right)\left(4x-1\right)+\left(3x+7\right)\left(-16x\right)=164\\ \left(48x^2-12x-20x+5\right)+\left(-48x^2-112x\right)=164\\ 48x^2-32x+5-48x^2-112x=164\\ -144x=164-5\\ -144x=159\\ x=-\dfrac{159}{144}\)

Check lại đề nha 

a. \(\left(5x+y\right)^2=25x^2+10xy+y^2\)

b. \(\left(3x-y\right)^2=9x^2-6xy+y^2\)

c. \(\left(\dfrac{1}{2}-3y\right)\left(\dfrac{1}{2}+3y\right)=\left[\left(\dfrac{1}{2}\right)^2-\left(3y\right)^2\right]\\ =\dfrac{1}{4}-9y^2\)

d. \(\left(2x+\dfrac{1}{3}y\right)^2=4x^2+\dfrac{4}{3}xy+\dfrac{1}{9}y^2\)

e. \(\left(\dfrac{1}{3}-y\right)^3=\left(\dfrac{1}{3}\right)^3-3\left(\dfrac{1}{3}\right)^2y+3\dfrac{1}{3}y^2-y^3\\=\dfrac{1}{27}-\dfrac{ 1}{9}y+y^2-y^3\)

f. \(\left(y-3\right)\left(y^2+3y+9\right)=y^3-27\)

\(\dfrac{3}{7}-x=\dfrac{1}{4}-\dfrac{-3}{5}\\ \dfrac{3}{7}-x=\dfrac{17}{20}\\ x=\dfrac{3}{7}-\dfrac{17}{20}\\ x=-\dfrac{59}{140}\)

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\(\left|x-\dfrac{3}{5}\right|-\dfrac{3}{4}=-\dfrac{1}{2}\\ \left|x-\dfrac{3}{5}\right|=-\dfrac{1}{2}+\dfrac{3}{4}\\ \left|x-\dfrac{3}{5}\right|=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{1}{4}\\x-\dfrac{3}{5}=-\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=-\dfrac{7}{20}\end{matrix}\right.\)

\(m_{Fe}=\dfrac{70.160}{100}=112\left(g\right)\)

\(m_O=160-112=48\left(g\right)\)

Số mol của mỗi nguyên tố là

\(n_{Fe}=\dfrac{112}{56}=2\left(mol\right)\)

\(n_O=\dfrac{48}{16}=3\left(mol\right)\)

Vậy CTHH : Fe₂O₃

PTK : 56 . 2 + 16 .3 = 160 ( DvC )