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

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

⇔ \(\frac{7}{6}-x=\frac{5}{6}+1\)

⇔ \(\frac{7}{6}-x=\frac{11}{6}\)

⇔ \(-x=\frac{11}{6}-\frac{7}{6}\)

⇔ \(-x=\frac{2}{3}\)

⇔ \(x=\frac{-2}{3}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{-2}{3}\right\}\)

A, \(\frac{x+3}{2}\)-\(\frac{x-1}{3}\)=\(\frac{x+5}{6}\)+1

⇔ \(\frac{3\left(x+3\right)}{6}\)-\(\frac{2\left(x-1\right)}{6}\)=\(\frac{x+5}{6}\)+\(\frac{6}{6}\)

⇔ 3x+9-2x+2=x+5+6

⇔ 3x-2x-x=5+6-9-2

⇔0x=0 (luôn đúng với mọi x)

Vậy phương trình có vô số nghiêm:S=R

bạn ktra lại đề nhé 

ĐKXĐ: x<>7

\(\frac{1}{7-x}=\frac{x-8}{x-7}-8\)

=>\(\frac{-1}{x-7}=\frac{x-8-8\left(x-7\right)}{x-7}\)

=>x-8-8(x-7)=-1

=>x-8-8x+56=-1

=>-7x+48=-1

=>-7x=-49

=>x=7(loại)

ĐK \(x\ge0\)

\(\Leftrightarrow\sqrt{x}+\sqrt{x+7}+x+2\sqrt{x\left(x+7\right)}+x+7=42\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)+\left(\sqrt{x}+\sqrt{x+7}\right)^2=42\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)^2+\left(\sqrt{x}+\sqrt{x+7}\right)-42=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{x+7}=6\\\sqrt{x}+\sqrt{x+7}=-7\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{x+7}\right)^2=36\)

\(\Leftrightarrow2x+7+2\sqrt{x\left(x+7\right)}=36\)

\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\)

bình phương 2 vế

\(\Leftrightarrow4\left(x^2+7x\right)=4x^2-116x+841\)

\(\Leftrightarrow4x^2+28x=4x^2-116x+841\)

\(\Leftrightarrow144x=841\Leftrightarrow x=\dfrac{841}{144}\)

ĐKXĐ: x<>2; x<>-2

Ta có: \(\frac{120}{x-2}-\frac{96}{x+2}=2\)

=>\(\frac{60}{x-2}-\frac{48}{x+2}=1\)

=>\(\frac{60x+120-48x+96}{\left(x-2\right)\left(x+2\right)}=1\)

=>(x-2)(x+2)=12x+216

=>\(x^2-4-12x-216=0\)

=>\(x^2-12x-220=0\)

=>(x-22)(x+10)=0

=>x=22(nhận) hoặc x=-10(nhận)

\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

... rồi đặt x+100 làm nhân tử chung => x = -100

trả lời đầy đủ cho mình với

ĐK : \(\hept{\begin{cases}x-7\ne0\\7-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne7\\x\ne7\end{cases}}\Leftrightarrow x\ne7\)

Ta có : \(\frac{x-8}{x-7}=\frac{1}{7-x}+8\Leftrightarrow\frac{x-8}{x-7}+\frac{1}{x-7}=8\Leftrightarrow\frac{x-8+1}{x-7}=8\)

\(\Rightarrow1=8\)(có mâu thuẫn) 

Vậy phương trình vô nghiệm 

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

\(\Rightarrow\frac{x-8}{x-7}=\frac{-1}{x-7}+\frac{8\left(x-7\right)}{x-7}\)

\(\Rightarrow x-8=8x-56-1\)

\(\Leftrightarrow x-8x=-57+8\)

\(\Leftrightarrow-7x=-49\)

\(\Leftrightarrow x=7\)

<=> \(\frac{2x-3}{35}\)+ \(\frac{x^2-2x}{7}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> \(\frac{2x-3}{35}\)+ \(\frac{5\left(x^2-2x\right)}{35}\)< \(\frac{5x^2}{35}\)- \(\frac{7\left(2x-5\right)}{35}\)

<=> 2x - 3 + 5( x² - 2x ) < 5x² - 7( 2x - 5 )

<=> 2x - 3 + 5x² - 10x < 5x² - 14x + 35

<=> 2x + 5x² - 10x - 5x² + 14x < 35 + 3

<=> 6x < 38

<=> x < \(\frac{19}{3}\)

có sai thì thông cảm :)