\(=5\sqrt{2^2\left[\left(x-4\right)^3\right]^2}-3\left(x-4\right)^3\)

\(=10\left|\left(x-4\right)^3\right|-3\left(x-4\right)^3\)

\(=-10\left(x-4\right)^3-3\left(x-4\right)^3\)

\(=-13\left(x-4\right)^3\)

Có: \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\forall x\\\left|y-1\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left|x-3\right|+\left|y-1\right|\ge0\forall x;y\)

Mà: \(\left|x-3\right|+\left|y-1\right|=0\)

nên: \(\left\{{}\begin{matrix}x-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

KO CẦN ĐỌC TIẾP NX

đag thi à?

Bài 4:

Điện trở tương đương :

\(R_{tđ}=\dfrac{U}{V}=\dfrac{12}{0,4}=30\left(\Omega\right)\)

Ta có: \(R_{tđ}=R_1+R_2\Leftrightarrow R_1=R_{tđ}-R_2=30-20=10\left(\Omega\right)\)

A B M N R1 R2 R3

Vì R > R_tđ ⇒ R₁ // R₂ \(\Rightarrow R_{tđ\left(MN\right)}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{30.30}{30+30}=15\left(\Omega\right)\)

Vì R_MN > R_tđ ⇒ R_MN // R₃  \(\Rightarrow R_{tđ\left(AB\right)}=\dfrac{R_{MN}.R_3}{R_{MN}+R_3}=\dfrac{15.30}{15+30}=10\left(\Omega\right)\)

Đó nha, khi qua ko đọc kĩ

Mà cái này có nhiều cách vẽ lắm

dấu . là nhân hay là phần ngăn cách ở hàng phần nghìn thế

a) \(28400:\left(100\times311-310\times100\right)\)
\(=28400:\left[\left(311-310\right)\times100\right]\)
\(=28400:100\)
\(=284\)
b) \(\dfrac{5}{11}+\dfrac{19}{15}+\dfrac{6}{11}+\dfrac{11}{16}+\dfrac{6}{15}+\dfrac{5}{16}\)
\(=\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+\left(\dfrac{19}{15}+\dfrac{6}{15}\right)+\left(\dfrac{11}{16}+\dfrac{5}{16}\right)\)
\(=1+\dfrac{5}{3}+1\)
\(=\dfrac{8}{3}+1\)
\(=\dfrac{11}{3}\)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

\(MgO+CO->\left(CO.ko.khử,đc\right)\)

\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Từ các pthh trên thấy: \(n_{CO_2.sinh.ra}=n_{CO.pứ}=0,2\left(mol\right)\left(theo.tỉ.lệ.pthh\right)\)

Áp dụng ĐLBTKL có: \(m_{hh}+m_{CO}=m_{rắn}+mCO_2\)

=> \(m_{rắn}=m_{hh}+m_{CO}-m_{CO_2}=12,5+0,2.28-0,2.44=9,3\left(g\right)\)