a) \(=7\left(x^2-3x+2\right)=7\left[x\left(x-1\right)-2\left(x-1\right)\right]=7\left(x-1\right)\left(x-2\right)\)

b) \(=2\left(x^2-4x+3\right)=2\left[x\left(x-3\right)-\left(x-3\right)\right]=2\left(x-3\right)\left(x-1\right)\)

c) \(=x\left(x-4\right)-\left(x-4\right)=\left(x-4\right)\left(x-1\right)\)

\(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{15.35}{15+35}=10,5\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{23}=20+10,5=30,5\left(\Omega\right)\)

\(I=I_1=I_{23}=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{36,6}{30,5}=1,2\left(A\right)\)

\(U_1=I_1.R_1=1,2.20=24\left(V\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

\(MgO+H_2SO_4\rightarrow MgSO_4\downarrow+H_2O\)

\(Al\left(OH\right)_3+3HNO_3\rightarrow Al\left(NO_3\right)_3+3H_2O\)

\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Gọi số HS lớp 7A,7B,7C lần lượt là a,b,c(HS)(a,b,c∈N*,a,b,c<144)

Ta có: \(\left(1-\dfrac{1}{4}\right)a=\left(1-\dfrac{1}{7}\right)b=\left(1-\dfrac{1}{3}\right)c\)

\(\Rightarrow\dfrac{3}{4}a=\dfrac{6}{7}b=\dfrac{2}{3}c\Rightarrow\dfrac{a}{\dfrac{4}{3}}=\dfrac{b}{\dfrac{7}{6}}=\dfrac{c}{\dfrac{3}{2}}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{\dfrac{4}{3}}=\dfrac{b}{\dfrac{7}{6}}=\dfrac{c}{\dfrac{3}{2}}=\dfrac{a+b+c}{\dfrac{4}{3}+\dfrac{7}{6}+\dfrac{3}{2}}=\dfrac{144}{4}=36\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{4}{3}.36=48\\b=\dfrac{7}{6}.36=42\\c=\dfrac{3}{2}.36=54\end{matrix}\right.\)(nhận)

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Đề bảo tìm j v bn?

\(\Rightarrow\left(x-7\right)\left(x^2-x+6\right)-x\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x^2-2x+6\right)=0\)

\(\Rightarrow x=7\left(do.x^2-2x+6=\left(x-1\right)^2+5\ge5>0\right)\)

\(S=\pi\dfrac{d^2}{4}=\pi\dfrac{0,5^2}{4}=\dfrac{157}{800}\left(mm^2\right)\)

\(R=p.\dfrac{l}{S}=1,1.10^{-6}.\dfrac{6}{\dfrac{157}{800}.10^{-6}}\approx33,6\left(\Omega\right)\)

Gọi số HS lớp 7A,7B,7C lần lượt là a,b,c(HS)(a,b,c∈N*)

Ta có: \(\dfrac{2a}{3}=b=\dfrac{4c}{5}\Rightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{5}{4}}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b-c}{\dfrac{3}{2}+1-\dfrac{5}{4}}=\dfrac{55}{\dfrac{5}{4}}=44\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}.44=66\\b=1.44=44\\c=\dfrac{5}{4}.44=55\end{matrix}\right.\)(Nhận)

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