|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

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

\(\frac{1}{x+1+\frac{1}{x}}+\frac{2}{x+2+\frac{1}{x}}=\frac{8}{15}\)

Đặt \(x+1+\frac{1}{x}=a\)

\(\frac{1}{a}+\frac{2}{a+1}=\frac{8}{15}\)

\(\Leftrightarrow a+1+2a=\frac{8}{15}a\left(a+1\right)\)

\(\Leftrightarrow8a^2-37a-15=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-\frac{3}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1+\frac{1}{x}=5\\x+1+\frac{1}{x}=-\frac{3}{8}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+1=0\\x^2+\frac{11}{8}x+1=0\end{matrix}\right.\)

\(\frac{2x-8}{6}-\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)

\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x+1\right)}{24}=\frac{3\left(9x-2\right)}{24}+\frac{2\left(3x-1\right)}{24}\)

\(\Leftrightarrow\frac{8x-32}{24}-\frac{18x+6}{24}=\frac{27x-6}{24}+\frac{6x-2}{24}\)

\(\Leftrightarrow8x-32-18x-6=27x-6+6x-2\)

\(\Leftrightarrow8x-18x-27x-6x=-6-2+32+6\)

\(\Leftrightarrow-42x=30\)

\(\Leftrightarrow x=-\frac{5}{7}\)

\(\hept{\begin{cases}\frac{7}{x-y+2}-\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{cases}}\)

Đặt \(a=\frac{1}{x-y+2};b=\frac{1}{x+y-1}\)ta được hệ phương trình:

\(\hept{\begin{cases}7a-5b=\frac{9}{2}\\3a+2b=4\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=\frac{1}{2}\end{cases}}}\)

Với \(\hept{\begin{cases}a=1\\b=\frac{1}{2}\end{cases}}\), ta được:

\(\hept{\begin{cases}\frac{1}{x-y+2}=1\\\frac{1}{x+y-1}=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y+2=1\\x+y-1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Vậy hệ phương trình có 1 nghiệm là x = 1 và y = 2 

a) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\)\(\frac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\frac{35\left(5x+4\right)+315}{105}\)

\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)

\(\Leftrightarrow84x+63-90x+30=175x+140+315\)

\(\Leftrightarrow84x-90x-175x=140+315-63-30\)

\(\Leftrightarrow-181x=362\)

\(\Leftrightarrow x=-2\)

b)\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x+4\right)^2}{6}=0\)

\(\Leftrightarrow\)\(\frac{8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x+4\right)^2}{24}=0\)

\(\Leftrightarrow8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2+8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2+32x+64=0\)

\(\Leftrightarrow8x^2-12x^2+4x^2-32x+32x=-64-27-32\)

\(\Leftrightarrow0x=-123\) (vô nghiệm)

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