A= E387E4837

B = 883433

C = UỲUWFHQWURY48E3947

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

            \(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)

Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)

\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)

1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)

Dấu "=" xảy ra khi x=y=1

Máy mình bị lỗi nên ko nhìn được các bài tiếp theo

Chúc bạn học tốt :)

Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2    

Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0

\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)

\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)

\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)

\(=>x=0\)

\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)

\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)

\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)

tu do \(=>x=-7,8;...;0;1;2;3;4\)
 

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Chắc đấy! Tick nhé!

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}

a)  \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)

Vậy....

b)  \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\)   vô lí do \(\left|a\right|\ge0\)

Vậy pt vô nghiệm

c)  \(\left|x+\frac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)

Vậy..

d)  \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do  \(\left|a\right|\ge0\)với mọi a

Vậy pt vô nghiệm

e)  \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)

\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)

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

Vậy...

pt là j v bn

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)