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: \(x=-\frac{5}{12}\)

https://hoc247.net/hoi-dap/toan-8/giai-phuong-trinh-2x-8x-1-2-4x-1-9-faq441870.html

\(\Rightarrow2x\cdot\left(64x^2-16x+1\right)\cdot\left(4x-1\right)=9\)

\(\Rightarrow\left(64x^2-16x+1\right)\cdot\left(8x^2-2x\right)=9\)

Nhân cả hai vế của phương trình với 8 ta được:

\(\left(64x^2-16x+1\right)\cdot\left(64x^2-16x\right)=72\)

Đặt \(a=64x^2-16x\left(a\ge1\right)\) (1)

\(\Rightarrow\left(a+1\right)\cdot a=72\)

\(\Rightarrow a^2+a-72=0\)

\(\Rightarrow\left(a-8\right)\cdot\left(a+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=8\left(tmđk\right)\\a=-9\left(loại\right)\end{matrix}\right.\)

Thay vào (1) ta đc:

\(64x^2-16x=8\Rightarrow64x^2-16x-8=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

$2x{}^{}\left(4x-1\right)=9$

$\iff \left(64{}^{}-16x+1\right)\left(64{}^{}-16x\right)=72$

Đặt 64x2 - 16x = t $\left(t\ge -1\right)$

$\Rightarrow t\left(t+1\right)=72$

$\iff \left(t+9\right)\left(t-8\right)=0$

⇔[ t=−9
t=8
⇔[t=−9(loai)t=8(nhan

$\Rightarrow 64{}^{}-16$

\(1.\sqrt{16-8x+x^2}=4-x\)

\(\sqrt{\left(4-x\right)^2}=4-x\)

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

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

\(2x-3-2x+3=0\)

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

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

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

a/ \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{\left(4-x\right)^2}=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\left|4-x\right|=4-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le4\\\left[{}\begin{matrix}4-x=4-x\left(loại\right)\\4-x=x-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

Vậy...

b/ \(\sqrt{4x^2-12x+9}=2x-3\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}2x-3=2x-3\left(loại\right)\\2x-3=3-2x\end{matrix}\right.\end{matrix}\right.\)

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

Vậy...

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

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

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

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

làm nốt

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

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

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

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

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

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

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......

Bài làm:

a) \(4x\left(x+2\right)=4x^2-24\)

\(\Leftrightarrow4x^2+8x=4x^2-24\)

\(\Leftrightarrow8x=-24\)

\(\Leftrightarrow x=-3\)

Vậy tập nghiệm của phương trình \(S=\left\{-3\right\}\)

b) \(\frac{x-2}{3}< \frac{8x-5}{9}\)

\(\Leftrightarrow\frac{3\left(x-2\right)}{9}< \frac{8x-5}{9}\)

\(\Leftrightarrow3x-6< 8x-5\)

\(\Leftrightarrow-5x< 1\)

\(\Leftrightarrow x>-\frac{1}{5}\)

Vậy \(x>-\frac{1}{5}\)

c) đkxđ: \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

Ta có: \(\frac{3}{x-2}+\frac{2}{x+2}=\frac{2x+5}{x^2-4}\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+5}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow3\left(x+2\right)+2\left(x-2\right)=2x+5\)

\(\Leftrightarrow3x+6+2x-4=2x+5\)

\(\Leftrightarrow3x=3\)

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

Vậy tập nghiệm của phương trình \(S=\left\{1\right\}\)

Học tốt!!!!