1a7 doanle bao anh 1234231 1234231 1234231

Bài 3 : 

Vì \(\left(x-2\right)^2\ge0\forall x\)

Nên :  \(A=\left(x-2\right)^2-4\ge-4\forall x\)

Vậy \(A_{min}=-4\) khi x = 2

B1: lấy máy tính mà tính thôi bạn (nhớ lm theo từng bước)

B2: 

a, \(\left|x-\frac{2}{3}\right|-\frac{1}{2}=\frac{5}{6}\)

\(\left|x-\frac{2}{3}\right|=\frac{4}{3}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{4}{3}\\x-\frac{2}{3}=\frac{-4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)

b, \(\frac{\left(-2\right)^x}{512}=-32\Rightarrow\left(-2\right)^x=-16384\Rightarrow x\in\varnothing\)

B3:

Vì \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x = 2

Vậy GTNN của A = -4 khi x = 2

Đề khó nhìn quá bạn ơi!

45^10*5^20/75^15

=5^10*9^10*5^20/(5^2)^15

=5^10*5^20*9^10/5^30

=9^10

(0.8)^5/(0.4)^6

=(0.4)^5*2^5/(0.4)^6

=2^5/(0.4)

=32/(0.4)

=80

2^15*9^4/6^6*8^3

=2^15*(3^2)^4/2^6*3^6*(2^3)^3

=2^15*3^8/2^6*3^6*2^9

=3^2

=9

Ta có: 

a) \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(5.3^2\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{5^{10}.3^{20}.5^{20}}{5^{30}.3^{15}}=\frac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)

b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2.2^2\right)^5}{\left(0,2.2\right)^6}=\frac{\left(0,2\right)^5.2^{10}}{\left(0,2\right)^6.2^6}=\frac{2^4}{0,2}=\frac{16}{0,2}=80\)

c) \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

=0 

Bài 1:

a)   \(x^2+5x=x\left(x+5\right)< 0\)  (1)

Nhận thấy:   \(x< x+5\)

nên từ (1)   \(\Rightarrow\)  \(\hept{\begin{cases}x< 0\\x+5>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< 0\\x>-5\end{cases}}\)\(\Leftrightarrow\)\(-5< x< 0\)

Vậy.....

b)   \(3\left(2x+3\right)\left(3x-5\right)< 0\)

TH1:   \(\hept{\begin{cases}2x+3>0\\3x-5< 0\end{cases}}\)\(\Leftrightarrow\)  \(\hept{\begin{cases}x>-\frac{3}{2}\\x< \frac{5}{3}\end{cases}}\)\(\Leftrightarrow\)\(-\frac{3}{2}< x< \frac{5}{3}\)

TH2:  \(\hept{\begin{cases}2x+3< 0\\3x-5>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< -\frac{3}{2}\\x>\frac{5}{3}\end{cases}}\)  vô lí

Vậy   \(-\frac{3}{2}< x< \frac{5}{3}\)

Bài 2:

a)  \(2y^2-4y=2y\left(y-2\right)>0\)

TH1:   \(\hept{\begin{cases}y>0\\y-2>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y>0\\y>2\end{cases}}\)\(\Leftrightarrow\)\(y>2\)

TH2:  \(\hept{\begin{cases}y< 0\\y-2< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y< 0\\y< 2\end{cases}}\)\(\Leftrightarrow\)\(y< 0\)

Vậy  \(\orbr{\begin{cases}y< 0\\y>2\end{cases}}\)

b)  \(5\left(3y+1\right)\left(4y-3\right)>0\)

TH1:  \(\hept{\begin{cases}3y+1>0\\4y-3>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y>-\frac{1}{3}\\y>\frac{3}{4}\end{cases}}\)\(\Leftrightarrow\)\(y>\frac{3}{4}\)

TH2:  \(\hept{\begin{cases}3y+1< 0\\4y-3< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y< -\frac{1}{3}\\y< \frac{3}{4}\end{cases}}\)\(\Leftrightarrow\)\(y< -\frac{1}{3}\)

Vậy   \(\orbr{\begin{cases}y>\frac{3}{4}\\y< -\frac{1}{3}\end{cases}}\)

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)