a. 9 - 35 = (17 - x) - (35 + 17)

=> -26 = 17 - x - 52

=> x = 17 - 52 + 26

=> x = -9

b. -7 - (12 + x) = -(-14 + 3) - (2x - 7)

=> -7 - 12 - x = - (-11) - 2x + 7

=> -19 - x = 11 + 7 - 2x

=> 2x - x = 18 + 19

=> x = 37

Thảo Jackson : nham

a) 3/35 - (3/5 + x) = 2/7

=> 3/5 + x= 3/35- 2/7

=> 3/5 +x = -1/5

=> x = -1/5 -3/5

=> x = -4/5

b) 3/7 +1/7 : x = 3/14

=> 1/7 : x= 3/14 -3/7

=> 1/7 : x = -3/14

=> x = 1/7 : -3/14 

=> x = -2/3

c) (5x-1).(2x-1/3)=0

=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)

Học tốt :D

a)x=-4/5

b)x=-2/3

c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy.........

mik lười mong bn thông cảm

a) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\\ \Rightarrow\dfrac{3}{5}+x=-\dfrac{1}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)

b) \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \Rightarrow\dfrac{1}{7}:x=-\dfrac{3}{14}\\ \Rightarrow x=-\dfrac{2}{3}\)

c) \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

\(460+85\times4=\left(x+200\right)\times4\)

\(\left(x+200\right)\times4=460+340\)

\(\left(x+200\right)\times4=800\)

\(x+200=800:4\)

\(x+200=200\)

\(x=200-200\)

\(x=0\)

~~~

\(\left(x-7\right)\left(2x-8\right)=0\)

\(+, TH1: x - 7 = 0\)

\(x=0+7\)

\(x=7\)

\(+, TH2 : 2x - 8 = 0 \)

\(2x=0+8\)

\(2x=8\)

\(x=8:2\)

\(x=4\)

~~~

\(x-280:35=5\times54\)

\(x-8=270\)

\(x=270+8\)

\(x=278\)

~~~

\(324+16\times\left(2x+3\right)=404\)

\(16\times\left(2x+3\right)=404-324\)

\(16\times\left(2x+3\right)=80\)

\(2x+3=80:16\)

\(2x+3=5\)

\(2x=5-3\)

\(2x=2\)

\(x=2:2\)

\(x=1\)

#\(Toru\)

`460 + 85 xx 4 = ( x + 200) xx 4`

`460 + 340 = (x+200)xx4`

` 800= (x+200)xx4`

`x+200=800:4`

`x+200=200`

`x=200-200`

`x=0`

__

`(x-7)(2x-8)=0`

`@ TH1`

`x-7=0`

`x=0+7`

`x=7`

`@ TH2`

`2x-8=0`

`2x=0+8`

`2x=8`

`x=8:2`

`x=4`

__

`x -280 : 35=5xx54`

`x -280 : 35=270`

`x-8=270`

`x=270+8`

`x=278`

__

`324 + 16xx(2x+3)=404`

`16xx(2x+3)=404 -324`

`16xx(2x+3)=80`

`2x+3=80:16`

`2x+3=5`

`2x=5-3`

`2x=2`

`x=2:2`

`x=1`

\(35-\left[\left(2x-3\right)^2:7\right]=28\)

\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)

\(\Rightarrow\left(2x-3\right)^2:7=7\)

\(\Rightarrow\left(2x-3\right)^2=1\)

\(\Rightarrow2x-3=\pm1\)

\(\Rightarrow x=2\) hay \(x=1\)

35 - [(2\(x\) - 3)²:7 ] = 28 

       (2\(x-3\))² : 7 = 35 - 28

       (2\(x\) - 3)² : 7 = 7

        (2\(x\) - 3)²      = 7 \(\times\) 7 

        (2\(x-3\))² = 7²

        \(\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\)

        \(\left[{}\begin{matrix}2x=-7+3\\2x=7+3\end{matrix}\right.\)

        \(\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

        Vậy \(x\in\) {-2; 5}

x ϵ {-2; 5}

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

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

\(\Leftrightarrow x=5\)

b,x²-2x-35=0

=>(x²-2x+1)-36=0

=>(x-1)²-6²=0

=>(x-1-6)(x-1+6)=0

=>(x-7)(x+5)=0

=>x=7 hoặc x=-5