\(B=5.3^2+4.3^2=3^2\left(5+4\right)=9.9=9^2\)

\(C=5^3+6^3+59=125+216+59=400=20^2\)

\(D=5.4^3+2^4.5+41=320+80+41=441=21^2\)

b: \(5\cdot3^2+4\cdot3^2=3^2\cdot9=3^4\)

c: \(5^3+6^3+59=20^2\)

\(5\cdot4^3+2^4\cdot5+41=21^2\)

\(\dfrac{2x-1}{5}-\dfrac{4x}{3}=2x-\dfrac{x}{10}\\ \Leftrightarrow\dfrac{6\left(2x-1\right)}{30}-\dfrac{40x}{30}=\dfrac{60x}{30}-\dfrac{3x}{30}\\ \Leftrightarrow12x-6-40x=60x-3x\\ \Leftrightarrow-28x-6=57x\\ \Leftrightarrow57x+28x+6=0\\ \Leftrightarrow85x=-6\\ \Leftrightarrow x=-\dfrac{6}{85}\)

ĐKXĐ : \(1\le x\le3\)

Ta có \(\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}\ge x^3+10\)

<=> \(-2\sqrt{x-1}-2\sqrt{3-x}-8x\sqrt{2x}\le-2x^3-20\)

<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+2x^3-8x\sqrt{2x}+16\le0\)(1)

Đặt \(\sqrt{2x}=y\) => \(x=\dfrac{y^2}{2}\)

Khi đó \(2x^3-8x\sqrt{2x}+16=\dfrac{y^6}{4}-4y^3+16=\left(\dfrac{y^3-8}{2}\right)^2\)

Khi đó (1) <=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\le0\)(1)

mà \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\ge0\forall x;y\)(2) 

Từ (2)(1) => \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2=0\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{3-x}-1=0\\\dfrac{y^3-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\3-x=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\\\sqrt{2x}=2\end{matrix}\right.\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm bất phương trình

\(a,3^{x+2}=7\\ \Leftrightarrow x+2=log_37\\ \Leftrightarrow x=log_37-2\approx-0.229\)

\(b,3\cdot10^{2x+1}=5\\ \Leftrightarrow10^{2x+1}=\dfrac{5}{3}\\ \Leftrightarrow2x+1=log\left(\dfrac{5}{3}\right)\\ \Leftrightarrow2x=log\left(\dfrac{5}{3}\right)-1\\ \Leftrightarrow x=\dfrac{1}{2}\cdot log\dfrac{5}{3}-\dfrac{1}{2}\\ \Leftrightarrow x\approx-0,389\)

Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

giúp mình nhé

a.\(\left|x-3\right|=4x+1\)

 \(ĐK:4x+1\ge0\Leftrightarrow4x\ge-1\Leftrightarrow x\ge\dfrac{-1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4x+1\\x-3=-4x-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4x=1+3\\x+4x=-1+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x=4\\5x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4}{3}\left(ktm\right)\\x=\dfrac{2}{5}\left(tm\right)\end{matrix}\right.\)

Vay S \(=\left\{\dfrac{2}{5}\right\}\)

b. \(\left|x-2\right|+2x=10\\ \Leftrightarrow\left|x-2\right|=10-2x\)

ĐK : \(10-2x\ge0\Leftrightarrow-2x\ge-10\Leftrightarrow x\le5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10-2x\\x-2=2x-10\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2x=10+2\\x-2x=-10+2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=12\\-x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=8\left(ktm\right)\end{matrix}\right.\)

Vay S \(=\left\{4\right\}\)