a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)

\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)

=>1/12x=77/57

=>x=308/19

b: =>(x^2-4)(x^2-10)=72

=>x^4-14x^2+40-72=0

=>x^4-14x^2-32=0

=>(x^2-16)(x^2+2)=0

=>x^2-16=0

=>x^2=16

=>x=4 hoặc x=-4

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

ĐKXĐ: ...

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

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\left(x+2\right)\left(x-3\right)+3=\left(x-4\right)\left(x+2\right)-7\)

\(\Leftrightarrow x^2-x-6+3=x^2-2x-8-7\)

\(\Leftrightarrow x^2-x-x^2+2x=6-3-8-7\)

\(\Leftrightarrow x=-12\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-12\right\}\)

\(x^2-3x+2x-6+3-x^2-2x+4x+8+7=0\)

\(x+12=0\)

\(x=-12\)

Lời giải:
ĐK: \(-2\leq x\leq 4\)

Ta có: \(x^2-2x+8-4\sqrt{(4-x)(x+2)}=0\)

\(\Leftrightarrow x^2-2x+8-4\sqrt{2x+8-x^2}=0\)

\(\Leftrightarrow 16-(2x-x^2+8)-4\sqrt{2x+8-x^2}=0\)

Đặt \(\sqrt{2x+8-x^2}=t\)

\(\Rightarrow 16-t^2-4t=0\)

\(\Rightarrow t=-2\pm 2\sqrt{5}\). Vì \(t\geq 0\Rightarrow t=-2+2\sqrt{5}\)

\(\Rightarrow t^2=2x+8-x^2=24-8\sqrt{5}\)

\(\Leftrightarrow x^2-2x+16-8\sqrt{5}=0\)

\(\Rightarrow x=1\pm \sqrt{8\sqrt{5}-15}\) (đều thỏa mãn)

Vậy............

Giải:

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

ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)

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

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)

\(\Leftrightarrow x-4=x-2\)

\(\Leftrightarrow0x=2\)

Vậy ...

\(\left(x+4\right)^2\)nhấn lộn.mn giúp đỡ