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

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

\(\Leftrightarrow x=5\)

= 3

bài gì ?

3^3+x + 3^2+x + 3^1+x =360

3x +3^3 + 3^2 + 3^1 =360

3x +27+9+3 =360

3x +39 =360

3x =360-39

3x =321

3x = 107

cảm ơn Nguyen The Toan

1/2+ y/1=3/x

1/2+2y/2=3/x

1+2y/2=3/x

1+2y.x=2.3

vì 1+2y là lẻ suy ra thuộc ước lẻ của 6 thuộc -+1,-+3

nếu x=1, y= 0

nếu x=-1 thì y=-2

nếu x=3 thì y= 1

nếu x=-3 thì y= -2

mk ko chắc lắm, tóm lại cách làm là vậy đó

\(3\left(x-2\right)+4\left(x-5\right)=23\)

\(\Rightarrow3x-6+4x-20-23=0\)

\(\Rightarrow7x-49=0\)

\(\Rightarrow x=7\)

    3(x-2)+4(x-5)=23

<=>3x-6+4x-20=23

<=>7x-26=23

<=>7x=49

<=>x=7

Vậy x=7

3 ( x -2 ) + 4(x-5)= 23

=> 3x-6+4x-20   =23

=> 7x-26            =23

=>  7x                 =23+26=49

=>         x            =\(\dfrac{49}{7}\)=7

Vậy x=7.

\(a,5.\left(x+35\right)=515\)

\(x+35=103\)

\(x=103-35\)

\(x=68\)

\(b,12.x-33=3^2.3^3\)

\(12x-33=243\)

\(12x=276\)

\(x=23\)

\(d,12.\left(x-1\right):3=4^3-2^3\)

\(12.\left(x-1\right):3=56\)

\(12.\left(x-1\right)=168\)

\(x-1=14\)

\(x=15\)

a) \(x^2-\left(x-3\right)\left(3x+1\right)=9\)

\(\Leftrightarrow x^2-9-\left(x-3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-3x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy nghiệm của pt x = 3 hoặc x = 1

Giải phương trình

a, x² - (x-3)(3x+1) = 9

\(\Leftrightarrow\) x² - 3x² + 8x +3 = 9

\(\Leftrightarrow\) -2x² + 8x - 6 = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, (x+14)³ - (x+12)³ =1352

\(\Leftrightarrow\) (x+14-x-12)[(x+14)² + (x+14)(x+12) + (x+12)² ] = 1352

\(\Leftrightarrow\) 6(x² + 28x + 196 + x² + 26x + 168 + x² +24x +144) =1352

\(\Leftrightarrow\) 18x² +468x + 3048 = 1352

Pt nghiệm vô tỉ