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

d: \(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)

=>\(\frac{x}{6}-\frac{1}{30}=\frac{2}{y}\)

=>\(\frac{5x-1}{30}=\frac{2}{y}\)

=>\(y\left(5x-1\right)=60\)

=>(5x-1;y)∈{(1;60);(60;1);(-1;-60);(-60;-1);(2;30);(30;2);(-2;-30);(-30;-2);(3;20);(20;3);(-3;-20);(-20;-3);(4;15);(15;4);(-4;-15);(-15;-4);(5;12);(12;5);(-5;-12);(-12;-5);(6;10);(10;6);(-6;-10);(-10;-6)}

=>(5x;y)∈{(2;60);(61;1);(0;-60);(-59;-1);(3;30);(31;2);(-1;-30);(-29;-2);(4;20);(21;3);(-2;-20);(-19;-3);(5;15);(16;4);(-3;-15);(-14;-4);(6;12);(13;5);(-4;-12);(-11;-5);(7;10);(11;6);(-5;-10);(-9;-6)}

=>(x;y)∈{(2/5;60);(61/5;1);(0;-60);(-59/5;-1);(3/5;30);(31/5;2);(-1/5;-30);(-29/5;-2);(4/5;20);(21/5;3);(-2/5;-20);(-19/5;-3);(1;15);(16/5;4);(-3/5;-15);(-14/5;-4);(6/5;12);(13/5;5);(-4/5;-12);(-11/5;-5);(7/5;10);(11/5;6);(-1;-10);(-9/5;-6)}

c: \(\frac{5}{x}-\frac{y}{3}=\frac16\)

=>\(\frac{5}{x}=\frac{y}{3}+\frac16=\frac{2y+1}{6}\)

=>x(2y+1)=30

=>(x;2y+1)∈{(1;30);(30;1);(-1;-30);(-30;-1);(2;15);(15;2);(-2;-15);(-15;-2);(3;10);(10;3);(-3;-10);(-10;-3);(5;6);(6;5);(-5;-6);(-6;-5)}

=>(x;2y)∈{(1;29);(30;0);(-1;-31);(-30;-2);(2;14);(15;1);(-2;-16);(-15;-3);(3;9);(10;2);(-3;-11);(-10;-4);(5;5);(6;4);(-5;-7);(-6;-6)}

=>(x;y)∈{(1;29/2);(30;0);(-1;-31/2);(-30;-1);(2;7);(15;1/2);(-2;-8);(-15;-3/2);(3;9/2);(10;1);(-3;-11/2);(-10;-2);(5;5/2);(6;2);(-5;-7/2);(-6;-3)}

b: \(\frac{4}{x}+\frac{y}{3}=\frac56\)

=>\(\frac{4}{x}=\frac56-\frac{y}{3}=\frac{5-2y}{6}\)

=>x(5-2y)=24

=>x(2y-5)=-24

=>(x;2y-5)∈{(1;-24);(-24;1);(-1;24);(24;-1);(2;-12);(-12;2);(-2;12);(12;-2);(3;-8);(-8;3);(-3;8);(8;-3);(4;-6);(-6;4);(-4;6);(6;-4)}

=>(x;2y)∈{(1;-19);(-24;6);(-1;29);(24;4);(2;-7);(-12;7);(-2;17);(12;3);(3;-3);(-8;8);(-3;13);(8;2);(4;-1);(-6;9);(-4;11);(6;1)}

=>(x;y)∈{(1;-19/2);(-24;3);(-1;29/2);(24;2);(2;-7/2);(-12;7/2);(-2;17/2);(12;3/2);(3;-3/2);(-8;4);(-3;13/2);(8;1);(4;-1/2);(-6;9/2);(-4;11/2);(6;1/2)}

a: \(\frac{x}{3}-\frac{4}{y}=\frac15\)

=>\(\frac{x}{3}-\frac15=\frac{4}{y}\)

=>\(\frac{5x-3}{15}=\frac{4}{y}\)

=>y(5x-3)=60

=>(5x-3;y)∈{(1;60);(60;1);(-1;-60);(-60;-1);(2;30);(30;2);(-2;-30);(-30;-2);(3;20);(20;3);(-3;-20);(-20;-3);(4;15);(15;4);(-4;-15);(-15;-4);(5;12);(12;5);(-5;-12);(-12;-5);(6;10);(10;6);(-6;-10);(-10;-6)}

=>(5x;y)∈{(4;60);(63;1);(2;-60);(-57;-1);(5;30);(33;2);(1;-30);(-27;-2);(6;20);(23;3);(0;-20);(-17;-3);(7;15);(18;4);(-1;-15);(-12;-4);(8;12);(15;5);(-2;-12);(-9;-5);(9;10);(13;6);(-3;-10);(-7;-6)}

=>(x;y)∈{(4/5;60);(63/5;1);(2/5;-60);(-57/5;-1);(1;30);(33/5;2);(1/5;-30);(-27/5;-2);(6/5;20);(23/5;3);(0;-20);(-17/5;-3);(7/5;15);(18/5;4);(-1/5;-15);(-12/5;-4);(8/5;12);(3;5);(-2/5;-12);(-9/5;-5);(9/5;10);(13/5;6);(-3/5;-10);(-7/5;-6)}

a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)

em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp

`a)1/3x-1/4x=1`

`(1/3-1/4)x=1`

`1/12x=1`

`x=1:1/12=12`

______________________________

`b)4/5+5/7:x=1/6`

`5/7:x=1/6-4/5`

`5/7:x=-19/30`

`x=5/7:(-19/30)`

`x=-150/133`

__________________________________

`c)[2x]/3-x/4=5/6`

`x(2/3-1/4)=5/6`

`x. 5/12=5/6`

`x=5/6:5/12=2`

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

1/12x = 1

x = 1 . 12

x = 12

b) 4/5 + 5/7: x = 1/6

5/7 : x = 1/6 - 4/5

5/7 : x = -19/30

x = 5/7 : -19/30

x = -150/133

c) 2x/3 - x/4 = 5/6

4 . 2x - 3x = 10

8x - 3x = 10

5x = 10

x = 10 : 5

x = 2.

a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0

a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran

Bài 1: Tính

a)  =1\(\frac{49}{60}\)

b)  =2\(\frac{13}{30}\)

Bài 2: Tìm x

a)  =4\(\frac{1}{21}\)

Riêng câu b) thì mk nghĩ là bạn viết lộn vì mk thấy cái chỗ xx3/4 là mk ko hiểu rồi