\(A=x^2-3x+5\)

\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)

Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)

a) \(A=x^2-3x+5\)

\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)

b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)

\("="\Leftrightarrow x=5\Rightarrow x=0;5\)

c) \(C=4x-x^2+3\)

\("="\Leftrightarrow x=7\Rightarrow x=2;7\)

d) \(D=x^4+x^2+2\)

\("="\Leftrightarrow x=2\Rightarrow x=0;2\)

2 x 2 x 2 x 1 x 2 x 2 x 3 x 4 x 7 x 5 x 2 x 0 = 0

\(2.2.2.1.2.2.3.4.7.5.2.0\)

\(=2^6.1.3.4.7.5.0\)

\(=0\)

~~~!!

a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)

\(=\left(\frac{3}{4}\right)^{45}:\left(\frac{3}{4}\right)^{20}\)

\(=\left(\frac{3}{4}\right)^{25}\)

b) \(\frac{125^{100}.2^{160}}{5^{298}.4^{80}}\)

\(=\frac{5^{300}.2^{160}}{5^{298}.2^{160}}\)

\(=5^2=25\)

a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)

\(\Leftrightarrow\frac{\left(\frac{3}{4}\right)^{45}}{\frac{3^{10}}{2^{10}}}=\frac{\frac{3^{45}}{4^{45}}}{\frac{3^{10}}{2^{10}}}=\frac{3^{45}.2^{10}}{4^{45}}=\frac{3^{35}.2^{10}}{2^{90}}=\frac{3^{35}}{2^{80}}\)

\(\Rightarrow\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}=\frac{3^{35}}{2^{80}}\)

x² - 25 - (x + 5) = 0

<=> (x - 5)(x + 5) - (x + 5) = 0

<=> (x + 5)( x - 5 - 1) = 0

<=> (x + 5)( x - 6) = 0 

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)

(2x - 1)² - (4x² - 1) = 0 

<=> (2x - 1)(2x - 1 - 2x - 1) = 0

<=> - 4x(2x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

Gọi 2 số nguyên đó là a, b ta có:

\(a^2+ab+b^2⋮9\) ta viết thành: \(\left(a-b\right)^2+3ab⋮9\Rightarrow\left(a-b\right)^2+3ab⋮3\)

Ta có: 

\(3ab⋮3\Rightarrow\left(a-b\right)^2⋮3\Rightarrow a-b⋮3\Rightarrow\left(a-b\right)^2⋮9\Rightarrow3ab⋮9\Rightarrow ab⋮3\)

ab chia hết cho 3 => có 1 số chia hết cho 3.
Mà a-b chia hết cho 3 nên 2 số có cùng số dư khi chia cho 3.
Vậy a,b chia hết cho 3 hay ab chia hết cho 9 (Q.E.D) => ĐPCM

Gọi 2 số lẻ liên tiếp là:   

2k−1và   

2k+1

Xét hiệu:    

A=(2k+1)^2−(2k−1)^2

=4k^2+4k+1−(4k^2−4k+1)

=8k ⋮8

⇒A⋮8

hay hiệu các bình phương của 2 số lẻ liên tiếp chia hết cho 8

\(E=5x^2+y^2+10+4xy-14x-6y\)

\(E=\left(2x+y-3\right)^2+\left(x-1\right)^2+6\)

Vì \(\left(2x+y-3\right)^2+\left(x-1\right)^2\ge0\)

Dấu '=" xảy ra.......................

C = \(y^2-2xy+x^2+2x^2-7\)

   = \(\left(y-x\right)^2+2x^2-7\)

Do \(\left(y-x\right)^2\ge0\)

      \(2x^2\ge0\)

=> \(\left(y-x\right)^2+2x^2-7\ge7\)

Min C = 7 <=> \(\hept{\begin{cases}2x^2=0=>x^2=0=>x=0\\y-x=0=>y=0\end{cases}}\)