Lời giải:

b.

$\frac{2x}{3}=8$

$\Leftrightarrow 2x=3.8=24$

$\Leftrightarrow x=24:2=12$
d.

$\frac{6}{5}x=-9$

$\Leftrightarrow x=-9: \frac{6}{5}=\frac{-15}{2}$

f.

$\frac{2-3x}{4}=\frac{4x-5}{5}$

$\Leftrightarrow 5(2-3x)=4(4x-5)$

$\Leftrightarrow 10-15x=16x-20$

$\Leftrightarrow 30=31x$

$\Leftrightarrow x=\frac{30}{31}$

h.

$\frac{10-3x}{2}=\frac{6x+1}{3}$

$\Leftrightarrow 3(10-3x)=2(6x+1)$

$\Leftrightarrow 30-9x=12x+2$

$\Leftrightarrow 28=21x$

$\Leftrightarrow x=\frac{28}{21}=\frac{4}{3}$

\(\dfrac{2x-3}{x-1}< \dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow6x-9< x-1\Leftrightarrow5x< 8\Leftrightarrow x< \dfrac{8}{5}\) và ĐK \(x\ne1\)

\(\dfrac{2x-3}{x-1}>\dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow x-1< 6x-9\Leftrightarrow5x>8\Leftrightarrow x>\dfrac{8}{5}\) và ĐK \(x\ne1\)

\(\Leftrightarrow\left(x+3\right)^2-48x^2=\left(x-3\right)^2\)

\(\Leftrightarrow48x^2=x^2+6x+9-x^2+6x-9\)

\(\Leftrightarrow48x^2-12x=0\)

=>12x(4x-1)=0

=>x=0(nhận) hoặc x=1/4(nhận)

\(-\dfrac{1}{2}x+6< 0\Leftrightarrow-\dfrac{1}{2}x< -6\Leftrightarrow\cdot\dfrac{1}{2}x>6\Leftrightarrow x>12\)

(sai thì thoi nha)

\(-\dfrac{1}{2}x+6< 0\)

\(\Leftrightarrow-\dfrac{1}{2}x< -6\)

\(\Leftrightarrow x>\left(-6\right):\left(-\dfrac{1}{2}\right)\)

\(\Leftrightarrow x>12\)

--> Chọn A

Tính tổng $S= 1^3 + 2^3 + 3^3 +....+ n^3$ - Các dạng toán khác - Diễn đàn Toán học

b: Ta có: \(\dfrac{12}{5}:x+\dfrac{4}{3}=3+\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{12}{5}:x=3-\dfrac{2}{3}=\dfrac{7}{3}\)

hay \(x=\dfrac{12}{5}:\dfrac{7}{3}=\dfrac{36}{35}\)

\(\left(\sqrt{3}-1\right)\cdot3+\dfrac{3\sqrt{3}}{2\sqrt{3}}=3\sqrt{3}-3+\dfrac{3}{2}=3\sqrt{3}-\dfrac{3}{2}\)