a: =>31-x=60

=>x=-29

b: =>(x-140):35=280-270=10

=>x-140=350

=>x=490

c: =>(1900-2x):35=48

=>1900-2x=1680

=>2x=220

=>x=110

d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)

=>2x-1=11

=>x=6

e: =>(x+2)^5=4^5

=>x+2=4

=>x=2

f: =>3x-4=0 hoặc x-1=0

=>x=4/3 hoặc x=1

g: =>(2x-1)^2=49

=>2x-1=7 hoặc 2x-1=-7

=>x=-3 hoặc x=4

h: =>x(x+1)/2=78

=>x(x+1)=156

=>x=12

a.

$xy=-21=7.(-3)=(-7).3=3.(-7)=(-3).7=21.(-1)=(-21).1=(-1).21=1(-21)$

Do đó $(x,y)=(7,-3); (-7,3); (3,-7); (-3,7); (21,-1); (-21,1); (-1,21); (1,-21)$

b.

$(x+5)(y-3)=14=1.14=14.1=(-14)(-1)=(-1)(-14)=2.7=7.2=(-2)(-7)=(-7)(-2)$

Do đó:

$(x+5,y-3)=(1,14); (14,1); (-14,-1); (-1,-14); (2,7); (7,2); (-2,-7); (-7,-2)$

Đến đây thì đơn giản rồi.

c.

$x(y-2)=-19$, bạn làm tương tự

d. Tương tự

Bài giải chi tiết đây em nhé:

\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+ \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

 \(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

      1   - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)

       1  -  \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)

               \(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)

                \(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)

                 \(2x+1\)  = 19

                 2\(x\)        = 19 - 1

                 2\(x\)       = 18

                    \(x\)      = 18: 2

                     \(x\)     = 9

Chọn đáp án C.

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

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

\(\Leftrightarrow x=5\)

trả lời thế thì ai mà chẳng trả lới đc