tách nhỏ ra

\(\dfrac{6}{8}-\dfrac{2}{7}=\dfrac{42}{56}-\dfrac{16}{56}=\dfrac{26}{56}=\dfrac{13}{28}\)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

\(x^2+4x+5=2\sqrt{2x+3}\)

\(ĐK:x\ge-\dfrac{3}{2}\)

\(pt\Leftrightarrow(2x+3-2\sqrt{2x+3}+1)+x^2+2x+1=0\)

\(\Leftrightarrow\left(\sqrt{2x+3}-1\right)^2=-\left(x+1\right)^2\)

Vì \(\left(\sqrt{2x+3}-1\right)^2\ge0;-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}(\sqrt{2x+3}-1)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}=1\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)}\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

\(\frac{\left(x-2\right)\left(2x-3\right)}{6}-\left(x+3\right)=\frac{x^2-12x}{3}\)

=>\(\frac{\left(x-2\right)\left(2x-3\right)-6\left(x+3\right)}{6}=\frac{2\left(x^2-12x\right)}{6}\)

=>(x-2)(2x-3)-6(x+3)=\(2\left(x^2-12x\right)\)

=>\(2x^2-3x-4x+6-6x-18=2x^2-24x\)

=>-13x-12=-24x

=>13x+12=24x

=>-11x=-12

=>x=12/11(nhận)

98

98

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)^2\)

\(y=\dfrac{23}{15}\)

23/15

a: \(B=\dfrac{x^2+5x+5x+25}{x\left(x+5\right)}=\dfrac{x+5}{x}\)

b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)

C

C