pt ko có VP sao giải đây

VP và 3,5 à bạn chờ nhé

\(\Leftrightarrow2\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=3,5\times2=7\)

\(\Leftrightarrow\left(8x+7\right)^2\left[2\left(4x+3\right)\right]\left[8\left(x+1\right)\right]=7\times8\)

Đặt t=8x+7.Pt trở thành:

\(t^2\left(t+1\right)\left(t-1\right)=56\)

\(\Leftrightarrow t^2\left(t^2-1\right)=56\)

\(\Leftrightarrow t^4-t^2-56=0\)

\(\Leftrightarrow t^4-8t^2+7t^2-56=0\)

\(\Leftrightarrow t^2\left(t^2-8\right)+7\left(t^2-8\right)=0\)

\(\Leftrightarrow\left(t^2-8\right)\left(t^2+7\right)=0\)

\(\Leftrightarrow t=8\)(vì t²+7>0)

Do đó 64x²+112x+41=0

Tới đây bạn denta hoặc vi-ét nó ra

\(\Leftrightarrow x=-\frac{\sqrt{2^3}+7}{8}\)hoặc\(x=\frac{\sqrt{2^3}-7}{8}\)

Đk: \(x\ge-2\)

PT \(\Leftrightarrow\) \(x\left(12-4\sqrt{x+2}\right)+3x^2-20x-7=0\)

\(\Leftrightarrow x.\dfrac{144-16\left(x+2\right)}{12+4\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\dfrac{-4x\left(x-7\right)}{3+\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left(3+\sqrt{x+2}\right)\left(3x+1\right)=4x\end{matrix}\right.\)

Đặt \(u=\sqrt{x+2}\Leftrightarrow x=u^2-2\left(u\ge0\right)\)

PT (2) \(\Leftrightarrow\left(3+u\right)\left(3u^2-5\right)=4\left(u^2-2\right)\)

\(\Leftrightarrow9u^2-15+3u^3-5u=4u^2-8\)

\(\Leftrightarrow3u^3+5u^2-5u-7=0\) \(\Leftrightarrow u=\dfrac{-1+\sqrt{22}}{3}\)

\(\Leftrightarrow x=\dfrac{5-2\sqrt{22}}{9}\)

Vậy...

Lời giải:
ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow 3x^2-20x-7=4x\sqrt{x+2}-12x$

$\Leftrightarrow (x-7)(3x+1)=4x(\sqrt{x+2}-3)=4x.\frac{x-7}{\sqrt{x+2}+3}$

$\Leftrightarrow x-7=0$ hoặc $3x+1=\frac{4x}{\sqrt{x+2}+3}$

Nếu $x-7=0\Leftrightarrow x=7$ (tm) 

Nếu $3x+1=\frac{4x}{\sqrt{x+2}+3}$

$\Leftrightarrow 9x+3+(3x+1)\sqrt{x+2}=4x$

$\Leftrightarrow 5x+3+(3x+1)\sqrt{x+2}=0$

$\Leftrightaqrrow 5x+3=-(3x+1)\sqrt{x+2}$

$\Rightarrow (5x+3)^2=(3x+1)^2(x+2)$

$\Leftrightarrow 9x^3-x^2-17x-7=0$

$\Leftrightarrow (x+1)(9x^2-10x-7)=0$

$\Rightarrow$........

\(a,x^2-10x-39=0\)

\(\Leftrightarrow x^2-10x-39+64=64\)

\(\Leftrightarrow x^2-10x+25=64\)

\(\Leftrightarrow\left(x-5\right)^2=64\)

làm nốt

\(x^2-10x-39=0\Leftrightarrow x^2-13x+3x-39=0\Leftrightarrow x\left(x-13\right)+3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)

\(b,\frac{x^2}{x^3-9}=\frac{1}{x+3}\)

\(\Leftrightarrow x^2\left(x+3\right)=x^3-9\)

\(\Leftrightarrow x^3+3x^2=x^3-9\)

\(\Leftrightarrow3x^2=-9\left(VL\right)\)

Đặt \(\sqrt{x^2+4x+7}=t>0\), ta có pt sau:

\(2\left(t^2+3\right)-7t=0\)

⇔ \(t^2-7t+6=0\Leftrightarrow\left(t-2\right)\left(2t-3\right)=0\)

⇔\(\left[{}\begin{matrix}t=2\\t=\frac{3}{2}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x^2+4x+7=4\\x^2+4x+7=\frac{9}{4}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\\x=\frac{\pm\sqrt{79}-4}{2}\end{matrix}\right.\)

Vậy ...

ĐKXĐ: \(x\ne2;x\ne1\)

Ta có: \(\frac{4x}{x-2}-\frac{1}{x-1}=\frac{8x^2-7}{3x-6}\)

\(\Leftrightarrow\frac{4x}{x-2}-\frac{1}{x-1}-\frac{8x^2-7}{3x-6}=0\)

\(\Leftrightarrow\frac{4x\left(x-1\right)\cdot3}{\left(x-2\right)\left(x-1\right)\cdot3}-\frac{1\left(x-2\right)\cdot3}{\left(x-1\right)\left(x-2\right)\cdot3}-\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}=0\)

\(\Leftrightarrow12x\left(x-1\right)-3\left(x-2\right)-\left(8x^2-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow12x^2-12x-\left(3x-6\right)-\left(8x^3-8x^2-7x+7\right)=0\)

\(\Leftrightarrow12x^2-12x-3x+6-8x^3+8x^2+7x-7=0\)

\(\Leftrightarrow-8x^3+20x^2-8x-1=0\)

\(\frac{12x\left(x-1\right)-3x+6}{3\left(x-2\right)\left(x-1\right)}=\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}\)

tiếp theo nhân vào và khử mẫu nha bạn!

mong mn giải ra kq lun xem

Trả lời:

\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)

\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)

\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)

\(\Leftrightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow10-3x=10-3x\)

\(\Leftrightarrow-3x+3x=10-10\)

\(\Leftrightarrow0x=0\)( luôn thỏa mãn )

Vậy S = R với \(x\ne0;x\ne2\)

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)