a) \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)

\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{x+1}{x-2}\Leftrightarrow A=x+1\)

b) \(\dfrac{M}{x-1}=\dfrac{x^2+3x+2}{x+1}\)

\(\Leftrightarrow\dfrac{M}{x-1}=\dfrac{\left(x+1\right)\left(x+2\right)}{x+1}\)

\(\Leftrightarrow\dfrac{M}{x-1}=x+2\Leftrightarrow M=\left(x-1\right)\left(x+2\right)=x^2+x-2\)

Câu 2:

\(R1=R_{nt}-R2=9-6=3\Omega\)

\(=>R_{ss}=\dfrac{R1\cdot R2}{R1+R2}=\dfrac{3\cdot6}{3+6}=2\Omega\)

Chọn A

nSO3=8/80=0,1(mol)

pthh: SO3 + H2O -> H2SO4

nH2SO4=nSO3=0,1(mol) => mH2SO4(tạo sau)= 0,1.98=9,8(g)

mH2SO4(tổng)= 100.9,8% + 9,8=19,6(g)

mddH2SO4(sau)=8+100=108(g)

=>C%ddH2SO4(sau)= (19,6/108).100=18,148%

tk:

a) \(\Rightarrow\left(x-3\right)\left(x+4\right)=5.12\)

\(\Rightarrow x^2+x-72=0\)

\(\Rightarrow\left(x-8\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

b) \(\Rightarrow\left(x+3\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-9\end{matrix}\right.\)

c) \(\Rightarrow2x^2=8\Rightarrow x^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Làm hết á!

b6, 

aTa có: |2x+4|≥0 với mọi x

⇒|2x+4|-5≥-5

Dấu "=" xảy ra ⇌ |2x+4|=0⇒x=-2

vậy GTNN P= -5 ⇌ x=-2

b, ta có (2x+7)\(^2\)≥0với mọi x

⇒(2x+7)\(^2\)+\(\dfrac{2}{5}\)≥\(\dfrac{2}{5}\)

Dấu "=" xảy ra⇌ x=-\(\dfrac{7}{2}\)

vậy GTNN E= \(\dfrac{2}{5}\)⇌x=-\(\dfrac{7}{2}\)

b7, ta có x\(^2\)≥0 với mọi x

⇒-x\(^2\)≤0 với mọi x

⇒-x\(^2\)-5≤-5

Dấu "=" xảy ra ⇌x=0

vậy GTLN A=-5⇌x=0