`c)1/4x+2/5=7/5`

`=>1/4x=7/5-1/5=1`

`=>x=1:1/4=4`

Vậy `x=4` 

`a)2x-2/3=-3/4`

`=>2x=-3/4+2/3=-1/12`

`=>x=-1/24`

Vậy `x=-1/24`

`b)x:3/4+1/4=-2/3`

`=>x:3/4=-2/3-1/4=-11/4`

`=>x=-11/4 xx 3/4=-33/16`

Vậy `x=-33/16`

\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)

\(< =>20x-5x^2+5x^2-12-x-6=0\)

\(< =>19x-18=0\)

\(< =>x=\dfrac{18}{19}\)

\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)

\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)

\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)

a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)

b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)

a)x=18/19

b)x=15/2

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

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

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a: \(\Leftrightarrow x^2-2x-8-x^2=36\)

=>-2x=44

hay x=-22

b: \(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)

=>-34x=3

hay x=-3/34

c: =>(x-10)(x-1)=0

=>x=10 hoặc x=1

a, Thay x=-3 vào A ta có:

\(A=2x^2-6x=2.\left(-3\right)^2-6.\left(-3\right)=2.9+6.3=18+18=39\)

Thay x=4 vào A ta có:

\(A=2x^2-6x=2.4^2-6.4=2.16-24=32-24=8\)

b, \(A=0\)

\(\Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

c, \(A=4x\)

\(\Leftrightarrow2x^2-6x=4x\\ \Leftrightarrow2x^2-10x=0\\ \Leftrightarrow2x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a, Thay x = -3 vào A ta được 

\(A=2.9-6.4=18-24=-6\)

Thay x = 4 vào A ta được 

\(A=2.16-6.4=32-24=6\)

b, Ta có \(A=2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\)

c, Ta có \(A=2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow x=0;x=5\)

a. \(x^2-2x+2\left|x-1\right|-7=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)

b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)

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

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

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)(x-6)^2-(x+6)^2=12`

`<=>(x-6-x-6)(x-6+x+6)=12`

`<=>-12.2x=12`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2`

`b)36x^2-12x+1=81`

`<=>(6x-1)^2=81`

`<=>(6x-1-9)(6x-1+9)=0`

`<=>(6x-10)(6x+8)=0`

`<=>(3x-5)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x-6)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2-6x+x-6=0`

`<=>x(x-6)+x-6=0`

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

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

a) 2 + 1/3 - x = 1 + 1/4

7/3 -x = 5/4

x = 7/3 - 5/4

x = 13/12

b) (2/7 x 2) : x = 1 :7/2 

4/7 : x = 2/7

x = 4/7 : 2/7

x = 2

a) 2 + 1/3 - x = 1 + 1/4

           7/3 -x = 5/4

                  x = 7/3 - 5/4

                  x = 13/12

b) (2/7 x 2) : x = 1 :7/2 

          4/7 : x = 2/7

                  x = 4/7 : 2/7

                  x = 2

a) 2 + 1/3 - x = 1 + 1/4

2 + 1/3 - x =  5/4

7/3 - x =  5/4 

x = 7/3 - 5/4

x = 13/12
b) (2/7 x 2) : x = 1 :7/2  

(2/7 x 2) : x = 2/7

4/7 : x = 2/7

x = 4/7 : 2/7

x = 2