\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)

\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}=9\)

\(\Leftrightarrow x+2=81\)

\(\Leftrightarrow x=79\)

\(\dfrac{2x+1}{3x+2}=\dfrac{x-1}{x-2}\) (đk: x≠ 2; \(-\dfrac{2}{3}\) )

⇔ \(\left(x-2\right)\left(2x+1\right)=\left(x-1\right)\left(3x+2\right)\)

⇔ \(2x^2+x-4x-2=3x^2+2x-3x-2\)

⇔ \(3x^2-x-2-2x^2+3x+2=0\)

⇔ \(x^2+2x=0\)

⇔ \(x\left(x+2\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;-2\right\}\)

\(\Leftrightarrow3x^2-3x+2x-2=2x^2-4x+x-2\)

\(\Leftrightarrow x^2+2x=0\)

=>x(x+2)=0

=>x=0 hoặc x=-2

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)

A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)

bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)

bài toán thành \(a^3+b^3+c^3-3abc=0\) nha

\(\Leftrightarrow\dfrac{5}{x^2+1}+\dfrac{7}{x^2+3}+\dfrac{9}{x^2+5}-\dfrac{4x^2+26}{x^2+10}=0\)

\(\Leftrightarrow\dfrac{5}{x^2+1}-1+\dfrac{7}{x^2+3}-1+\dfrac{9}{x^2+5}-1-\dfrac{4x^2+26}{x^2+10}+3=0\)

\(\Leftrightarrow\dfrac{4-x^2}{x^2+1}+\dfrac{4-x^2}{x^2+3}+\dfrac{4-x^2}{x^2+5}-\dfrac{x^2-4}{x^2+10}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\dfrac{1}{x^2+1}+\dfrac{1}{x^2+3}+\dfrac{1}{x^2+5}+\dfrac{1}{x^2+10}\right)=0\)

\(\Leftrightarrow4-x^2=0\)(vì \(\dfrac{1}{x^2+1}+\dfrac{1}{x^2+3}+\dfrac{1}{x^2+5}+\dfrac{1}{x^2+10}>0\))

\(\Leftrightarrow x=\pm2\)

Lời giải:

Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=1+(3+m)=4+m\geq 0\Leftrightarrow m\geq -4$ (chứ không phải với mọi m như đề bạn nhé)!

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=-2\\ x_1x_2=-(m+3)\end{matrix}\right.\)

$x_1, x_2\neq 0\Leftrightarrow -(m+3)\neq 0\Leftrightarrow m\neq -3$

$\frac{x_1}{x_2}-\frac{x_2}{x_1}=\frac{-8}{3}$

$\Leftrightarrow \frac{x_1^2-x_2^2}{x_1x_2}=\frac{-8}{3}$

$\Leftrightarrow \frac{-2(x_1-x_2)}{-(m+3)}=\frac{-8}{3}$
$\Leftrightarrow x_1-x_2=\frac{4}{3}(m+3)$

$\Rightarrow (x_1-x_2)^2=\frac{16}{9}(m+3)^2$

$\Leftrightarrow (x_1+x_2)^2-4x_1x_2=\frac{16}{9}(m+3)^2$
$\Leftrightarrow 4+4(m+3)=\frac{16}{9}(m+3)^2$

$\Leftrightarrow m+3=3$ hoặc $m+3=\frac{-3}{4}$

$\Leftrightarrow m=0$ hoặc $m=\frac{-15}{4}$ (đều thỏa mãn)

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(Đk:\) \(x\ne1,x\ne2,x\ne3\)

\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)

\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Rightarrow0x-14=x-10\)

\(\Rightarrow x=-4\left(tmđk\right)\)

1:

a: =>28x-8=9x+3

=>19x=11

=>x=11/19

b: =>(3x-1)(x-1)=(2x+1)(x+1)

=>3x^2-4x+1=2x^2+3x+1

=>x^2-7x=0

=>x=0 hoặc x=7