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

<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)

<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)

<=> \(-\frac{1}{3}x=\frac{29}{12}\)

<=> \(x=-\frac{29}{4}\)

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

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

<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)

<=> \(-\frac{1}{6}x=-\frac{5}{6}\)

<=> \(x=5\)

học tốt

1.

\(\Delta'=\left(m+1\right)^2-\left(m^2+3m-7\right)=-m+8\)

Phương trình đã cho vô nghiệm khi và chỉ khi \(\Delta'< 0\)

\(\Rightarrow-m+8< 0\)

\(\Rightarrow m>8\)

Bài 5: 

a: \(x\left(x-1\right)-x^2+4x=-3\)

\(\Leftrightarrow x^2-x-x^2+4x=-3\)

hay x=-1

i: \(x^2-9x+8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

1 + 1 = 2

Mn giúp mk với nha.

Mk tk lại cho.

Trả lời : Tk mk trước đi .

Hok_Tốt

#Thiên_Hy

1 + 1 = 2 nha !

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)