\(\frac{3}{7}\cdot15\cdot\frac{1}{3}+\frac{3}{7}\cdot5\cdot\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)

\(\Leftrightarrow\frac{15}{7}+\frac{6}{7}\le x\le-6\cdot\frac{-5}{3}\)

\(\Leftrightarrow3\le x\le10\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)

bn Quân sai rồi, hỗn số \(15\frac{1}{3}\)chứ có phải \(15.\frac{1}{3}\)đâu???

\(\frac{3}{7}\cdot15\frac{1}{3}+\frac{3}{7}\cdot5\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)

\(\Rightarrow\frac{3}{7}\left(15\frac{1}{3}+5\frac{2}{5}\right)\le x\le\left(\frac{7}{2}:7-\frac{13}{2}\right)\cdot\left(-\frac{7}{3}\right)\)

\(\Rightarrow\frac{3}{7}\left(\frac{46}{3}+\frac{27}{5}\right)\le x\le\left(\frac{7}{14}-\frac{13}{2}\right)\cdot\left(-\frac{7}{3}\right)\)

\(\Rightarrow\frac{311}{35}\le x\le14\)

\(\Rightarrow8,8...\le x\le14\)

Đến đây là tìm x :))

x=2+3=5

\(\frac{2^x}{5^x}>\frac{2^3}{5^3}.\frac{\left(-2\right)^2}{5^2}\)

\(\frac{2^x}{5^x}>\frac{2^5}{5^5}\)

\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)

\(\Rightarrow x>5\)

vậy \(x>5\)

\(\frac{2^{4-x}}{16^5}=32^6\)

=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)

=> \(2^{4-x}=2^{30}.2^{20}\)

=> \(2^{4-x}=2^{50}\)

=> 4  - x = 50

=> x = 4 - 50 = -46

\(\frac{3^{2x+3}}{9^3}=9^{14}\)

=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)

=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)

=> \(3^{2x+3}=3^{28}.3^6\)

=> \(3^{2x+3}=3^{34}\)

=> 2x + 3 = 34

=> 2x = 34 - 3

=> 2x = 31

=> x = 31/2

ta có : \(x\ne3\) để mẫu khác 0

Vì 2 phân số có cùng mẫu nên

\(\left|x-5\right|=\left|x-1\right|\)

*TH1: \(\begin{cases}x-5\ge0\\x-1\ge0\end{cases}\)

\(x-5=x-1\)

\(0x=4\)

KHông có giá trị x

*TH2:

\(\begin{cases}x-5\le0\\x-1\le0\end{cases}\)

\(-\left(x-5\right)=-\left(x-1\right)\)

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

\(0x=-4\)

Không có giá trị x

*TH3:

\(\begin{cases}x-1\ge0\\x-5\le0\end{cases}\) \(\Rightarrow\begin{cases}x\ge1\\x\le5\end{cases}\)

\(-\left(x-5\right)=x-1\)

\(\Rightarrow5+1=2x\)

\(\frac{6}{2}=x\)

\(x=3\)

Mà \(x\ne3\) 

nên ko có giá trị thỏa mãn

vậy không có giá trị x nguyên thỏa mãn với đề bài

x=\(-\frac{125}{27}\)

\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)

\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)

\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)

TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)

CHUC BAN HOC TOT :))