Số ban đầu \(\overline{ab}=10\times a+b\)

Số lúc sau : \(\overline{a0b}=100\times a+b\)

Số lúc sau gấp 6 lần số ban đầu :

\(\left(10\times a+b\right)\times6=100\times a+b\)

\(\Leftrightarrow60\times a+6\times b=100\times a+b\)

\(\Leftrightarrow40\times a-6\times b=0\)

\(\Leftrightarrow20a-3b=0\)

\(\Leftrightarrow20a=3b\Leftrightarrow a=\frac{3b}{20}\)

Mỗi quyển vở có giá là :

\(22500:15=a\)( đồng )

Vậy 9000 đồng mua đước số quyển là :

9000 : a =...........

\(16x^2+y^2+4y-16x-8xy\)

\(=\left(4x-y\right)^2-4\left(4x-y\right)\)

\(=\left(4x-y\right)\left(4x-y-4\right)\)

\(A=-4,5+\left|x-1,5\right|\ge-4,5\)

\(Min_A=-4,5\)

\(\Leftrightarrow x=1,5\)

làm mẫu câu a nha :

\(\left|x-3\right|+1=3\)

xét \(x-3\ge0\)

\(\Leftrightarrow x-3+1=3\Rightarrow x=5\)

Xét \(x-3< 0\)

\(\Leftrightarrow3-x+1=3\)

\(\Rightarrow x=1\)

\(\left(3x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

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

\(=\left|1-3x\right|+\left|3x-2\right|\)

\(\ge\left|1-3x+3x-2\right|=\left|-1\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(1-3x\right)\left(3x-2\right)\ge0\Leftrightarrow\frac{1}{3}\le x\le\frac{2}{3}\)

Vậy \(A_{min}=1\) tại \(\frac{1}{3}\le x\le\frac{2}{3}\)

\(A=2\left|x-5\right|-2015\ge-2015\)

\(Min_A=-2015\Leftrightarrow x=5\)

\(B=205-\left|3x-5\right|\le205\)

\(Max_B=205\Leftrightarrow x=\frac{5}{3}\)

\(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x+41=1\)

\(\Leftrightarrow2x=42\)

\(\Leftrightarrow x=21\)

\(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x+41=1\)

\(\Leftrightarrow2x=42\)

\(\Leftrightarrow x=21\)