b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

a, x=-2/3

b, x=-5/7

c, x=11

chắc chắn đúng luôn

a, x=-2/3

b, x=-5/7

c, x=11

Đây là toán lớp 7 mà bạn.

BBạn có trả lời dc ko, ko thì thôi

a) |3x-2|=7

=> 3x-2=7 hoặc 3x-2=-7

=> 3x=9 hoặc 3x=-5

=> x=3 hoặc x=\(-\dfrac{5}{3}\)

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

Ko chép đề

\(2)x-17+x=x-7\)

\(x+x-x=-7+17\)

\(x=-10\)

Vậy ....

âm 10 , nhỉ ?

1a) \(\frac{x-3}{x+7}=\frac{-5}{-6}\)

=> \(\frac{x-3}{x+7}=\frac{5}{6}\)

=> (x - 3).6 = 5.(x + 7)

=> 6x - 18 = 5x + 35

=> 6x - 5x = 35 + 18

=> x = 53

b) \(\frac{x-7}{x+3}=\frac{4}{3}\)

=> (x - 7). 3 = (x + 3). 4

=> 3x - 21 = 4x + 12

=> 3x - 4x = 12 + 21

=> -x = 33

=> x = -33

c) \(\frac{x-10}{6}=-\frac{5}{18}\)

=> (x - 10) . 18 = -5 . 6

=> 18x - 180 = -30

=> 18x = -30 + 180

=> 18x = 150

=> x = 150 : 18 = 25/3

d) \(\frac{x-2}{4}=\frac{25}{x-2}\)

=> (x - 2)(x - 2) = 25 . 4

=> (x - 2)² = 100

=> (x - 2)² = 10²

=> \(\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}}\)

=> \(\orbr{\begin{cases}x=12\\x=-8\end{cases}}\)

e) \(\frac{7}{x}=\frac{x}{28}\)

=> 7 . 28 = x . x

=> 196 = x²

=> x² = 14²

=> \(\orbr{\begin{cases}x=14\\x=-14\end{cases}}\)

f) \(\frac{40+x}{77-x}=\frac{6}{7}\)

=> (40 + x) . 7 = (77 - x).6

=> 280 + 7x = 462 - 6x

=> 280 - 462 = -6x + 7x

=> -182 = x

=> x = -182

36367 l'290209 92021

\(a,\frac{3}{35}-\left[\frac{3}{5}+x\right]=\frac{2}{7}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2\cdot5}{35}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{10}{35}=\frac{-7}{35}\)

\(\Rightarrow x=\frac{-7}{35}-\frac{3}{5}=\frac{-7}{35}-\frac{3\cdot7}{35}=\frac{-7}{35}-\frac{21}{35}=\frac{-28}{35}=\frac{-4}{5}\)

\(b,\frac{5}{6}+\frac{1}{6}:x=\frac{3}{4}\)

\(\Rightarrow\frac{1}{6}:x=\frac{3}{4}-\frac{5}{6}\)

\(\Rightarrow\frac{1}{6}:x=\frac{18}{24}-\frac{20}{24}\)

\(\Rightarrow\frac{1}{6}:x=\frac{-2}{24}\)

\(\Rightarrow x=\frac{1}{6}:\frac{-2}{24}=\frac{1}{6}:\frac{-1}{12}=\frac{12}{-6}=\frac{-12}{6}=-2\)

\(c,\left|x-3\frac{4}{7}\right|=-0,5\)

Vì \(\left|x-3\frac{4}{7}\right|\ge0\)mà \(-0,5< 0\)nên \(x\in\varnothing\)

Vậy không có giá trị x thỏa mãn

a)  3/35-(3/5+x)=2/7

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

<=> 3/5+x=3/35-10/35

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

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

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

<=> x=-4/5

Vậy x=-4/5

(x-1/2).5/3=7/4-1/2

(x-1/2).5/3=5/4

x-1/2=5/4:5/3

x=3/4+1/2

x=5/4

(x-4/3).7/4=5-7/6

(x+4/3).7/4=23/6

x+4/3=23/6:7/4

x+4/3=46/21

x=46/21-4/3

x=6/7

a)x=\(\frac{5}{12}\)

b)x=\(\frac{39}{35}\)

c)-x=-\(\frac{4}{21}\)

x=\(\frac{4}{21}\)