Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

A,-8

B,-36

C,-80

{Mình tính bằng máy tính }

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

1)\(\left(x+1\right).\left(y-2\right)=0\)                                       \(\left(x,y\inℤ\right)\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)

2)\(\left(x-5\right).\left(y-7\right)=1\)

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

4)\(\left(x-4\right).\left(y+3\right)=-3\)

5)\(\left(x+3\right).\left(y-6\right)=-4\)

6)\(\left(x-8\right).\left(y+7\right)=5\)

7)\(\left(x+7\right).\left(y-3\right)=-6\)

8)\(\left(x-6\right).\left(y+2\right)=7\)

ok :)

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6