Ta có:

$\dfrac{x^4}{2x^2+1}+\dfrac{2x^2+1}{x^4}=2$

Điều kiện: $x\ne0$.

Đặt $t=\dfrac{x^4}{2x^2+1}>0$.

Khi đó:

$t+\dfrac{1}{t}=2$

$t^2-2t+1=0$

$(t-1)^2=0$

$t=1$

$\dfrac{x^4}{2x^2+1}=1$

$x^4=2x^2+1$

$x^4-2x^2-1=0$

Đặt $y=x^2\ge0$:

$y^2-2y-1=0$

$y=1+\sqrt2$ (do $y\ge0$)

$x^2=1+\sqrt2$

$x=\pm\sqrt{1+\sqrt2}$

Vậy $x=\pm\sqrt{1+\sqrt2}$.

Ta có:

$\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}$

Điều kiện: $x\ne\pm1$.

Quy đồng:

$\dfrac{x^2(x+1)^2+x^2(x-1)^2}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{x^2[(x+1)^2+(x-1)^2]}{(x^2-1)^2}=\dfrac{10}{9}$

$\dfrac{2x^2(x^2+1)}{(x^2-1)^2}=\dfrac{10}{9}$

$18x^2(x^2+1)=10(x^2-1)^2$

$9x^2(x^2+1)=5(x^2-1)^2$

$9x^4+9x^2=5x^4-10x^2+5$

$4x^4+19x^2-5=0$

Đặt $t=x^2\ge0$:

$4t^2+19t-5=0$

$(4t-1)(t+5)=0$

$t=\dfrac14$ (do $t\ge0$)

$x^2=\dfrac14$

$x=\pm\dfrac12$

Vậy $x=\pm\dfrac12$.

Ta có:

$x^3+3x^2-10x-24=0$

Nhóm các hạng tử:

$x^2(x+3)-10(x+3)=0$

$(x+3)(x^2-10)=0$

$x+3=0$ hoặc $x^2-10=0$

$x=-3$ hoặc $x=\pm\sqrt{10}$

Vậy $x\in{-3,-\sqrt{10},\sqrt{10}}$.

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

\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow x^2+30x+25=x^2+25x\)

=>5x=-25

hay x=-5(loại)

b: \(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)

=>2x+7=10

hay x=3/2

\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)

Quy đồng làm nốt

Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:

ĐKXĐ: ...

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)

Đặt \(4x+\frac{7}{x}-10=t\)

\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)

cảm ơn

1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

\(\Leftrightarrow\sqrt{x-2}=2\)

\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)

3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)