\(\sqrt{\sqrt{3}}.\sqrt{\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2}=\sqrt{3}\)

\(\Leftrightarrow\left(\dfrac{x-2}{28}-1\right)+\left(\dfrac{x-3}{27}-1\right)+\left(\dfrac{x-7}{23}-1\right)=0\)

\(\Leftrightarrow\left(x-30\right)\left(\dfrac{1}{28}+\dfrac{1}{27}+\dfrac{1}{23}\right)=0\)

\(\Leftrightarrow x-30=0\left(vì\left(\dfrac{1}{28}+\dfrac{1}{27}+\dfrac{1}{23}>0\right)\right)\)

\(\Leftrightarrow x=30\)

\(2^x=8.64=2^3.2^6=2^9\)

\(\Rightarrow x=9\)

\(4^x=4^{11}\Rightarrow x=11\)

\(7^{x+3}+7^x=2^3.2107\)

\(7^x\left(7^3+1\right)=2^3.2107\)

\(7^x.344=2^3.2107\)

\(7^x=49=7^2\)

\(x=2\)

a) \(-\dfrac{7}{12}\le x\le\dfrac{3}{12}\)

\(x\in\left\{-\dfrac{7}{12};-\dfrac{6}{12};...;\dfrac{3}{12}\right\}\)

b) \(-\dfrac{1}{12}< x< \dfrac{1}{8}\)

\(\Leftrightarrow-\dfrac{2}{24}< x< \dfrac{3}{24}\)

\(x\in\left\{-\dfrac{1}{24};...;\dfrac{2}{24}\right\}\)

Gọi 2 số chính phương đó là: \(a^2\) và \(b^2\)

Ta có:

\(n=a^2+b^2\)

\(n^2=a^4+2\left(ab\right)^2+b^4\)

\(n^2=\left(a^4-2\left(ab\right)^2+b^4\right)+4\left(ab\right)^2\)

\(n^2=\left(a^2-b^2\right)^2+\left(2ab\right)^2\left(đpcm\right)\)

anh giải câu b nhá, câu a anh không biết đề em viết kiểu sao

\(2^x+2^{x+3}=144\)

\(2^x+2^x.8=144\)

\(2^x\left(1+8\right)=144\)

\(2^x=16=2^4\)

\(\Rightarrow x=4\)

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\)

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

\(\Rightarrow\dfrac{2x}{15}=\dfrac{3y}{60}=\dfrac{z}{28}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{15}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{15+60+28}=\dfrac{186}{103}\)

đến đây em tự giải tiếp nha, câu b tương tự