1. A

2. D

\(97-\left(64-x\right)=44\)

\(64-x=97-44\)

\(64-x=53\)

\(x=64-53\)

\(x=11\)

I haven't met my grandma for years.

\(\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{3}{5}\)

Biểu thức luôn xác định là: \(\sqrt{x^2+x+1}\)

Giải thích: Vì biểu thức trong căn luôn > 0 với mọi số thực x.

\(7^{4x}+2=49^x+1\)

\(\Leftrightarrow4x+2=2\left(x+1\right)\)

\(\Leftrightarrow4x+2=2x+2\)

\(\Leftrightarrow4x-2x=2-2\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

\(\dfrac{a}{4}-\dfrac{2}{b}=\dfrac{5}{4}\)

\(\dfrac{ab-8}{4b}=\dfrac{5}{4}\)

\(\Rightarrow20b=4ab-32\)

\(4ab-20b=32\)

\(b\left(4a-20\right)=32\)

\(b=\dfrac{32}{4a-20}\)

\(b=\dfrac{4\cdot8}{4\left(a-5\right)}\)

\(b=\dfrac{8}{a-5}\)

a) Nhôm cacbua

b) Ancol benzylic

c) Natri etoxit

d) N,N-Đietyletanamin

e) But-2-in

g) Bạc axetylua

h) Propinylsilver

i) (1,2-Đibrometyl)benzen

ĐK: \(x\ge0\)

Lấy P - 1

\(\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}-1\)

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

\(=\dfrac{-\sqrt{x}-3}{2\sqrt{x}+1}\)

\(=\dfrac{-\left(\sqrt{x}+3\right)}{2\sqrt{x}+1}\)

Ta thấy \(\left\{{}\begin{matrix}\sqrt{x}+3>0\\2\sqrt{x}+1>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-\left(\sqrt{x}+3\right)< 0\\2\sqrt{x}+1>0\end{matrix}\right.\Rightarrow P-1< 0\)

Vậy \(P< 1\).

a) Khi x = 9 TMĐK thì: \(A=\dfrac{\sqrt{9}-2}{\sqrt{9}-2}=\dfrac{3-2}{3-1}=\dfrac{1}{2}\)

Vậy \(x=9\) thì \(A=\dfrac{1}{2}\).

b) Với \(x\ge0 ; x\ne 1\) thì:

\(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{5\sqrt{x}+3}{1-x}+\dfrac{4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{5\sqrt{x}+3}{x-1}+\dfrac{4}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-5\sqrt{x}-3+4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}-5\sqrt{x}-3+4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) (đpcm)