Câu d:

-1\(\frac23\) - (|2\(x\)| + \(\frac56\)) = - 2

-\(\frac53\) - |2\(x\)| - \(\frac56\) = - 2

|2\(x\)| = - \(\frac53\) - \(\frac56\) + 2

|2\(x\)| = - \(\frac52\) + 2

|2\(x\)| = - \(\frac12\) (vô lí vì trị tuyệt đối của một số luôn là một số không âm)

Không có giá trị nào của x thỏa mãn đề bài.

x ∈ ∅

Câu a:

|\(x\) - 3| = \(x\) + 4

Vì |\(x\) - 3| ≥ 0 ∀ \(x\) nên \(x\) + 4 ≥ 0 ⇒ \(x\) ≥ - 4

Với -4 ≤ \(x\) ≤ 3 ta có:

-\(x\) + 3 = \(x\) + 4

\(x\) + \(x\) = -4 + 3

2\(x\) = -1

\(x=\frac{-1}{2}\)

Với x > 3 ta có:

x - 3 = x + 4

x - x = 3 + 4

0 = 7 (vô lí)

Vậy x = -1/2 là nghiện duy nhất của phương trình.

Vậy \(x\) = -1/2

Bài 1 : 

\(A=x^2-2xy^2+y^4=\left(x-y^2\right)^2=-\left(y^2-x\right)^2\)

Mà \(B=-\left(y^2-x\right)^2\)

Nên ta có : đpcm 

Bài 2 

Đặt \(\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)

TH1 : x = -1

TH2 : x = 2

TH3 : x = 1/2 

Bài 4 : 

a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)

b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)

c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)

d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)

B1:

Ta có: a - b = ab => a = ab + b = b(a + 1)

Thay a = b(a + 1) vào a  - b  = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)

=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1 

Lại có: ab = a - b

<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2

Vậy...

B2:

a, \(3y\left(y-\frac{2}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)

b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)

\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)

B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)

\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)

\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)

\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha

a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)

=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)

=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)

b) \(2x-\left|x+1\right|=\frac{1}{2}\)

=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))

=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)

=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)

Câu c:

|2x -1| - |x + 1/3| = 0

|2x -1| = |x + 1/3|

2x - 1 = x + 1/3 hoặc 2x - 1 = - x + 1/3

2x - x = 1/3 + 1

x = 4/3

2x - 1 = - x + 1/3

2x + x = 1/3 + 1

3x = 4/3

x = 4/3 : 3

x = 4/9

Vậy x ∈ {4/9; 4/3}

\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)

\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)

\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)

\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)

\(\Leftrightarrow\text{ }x=\frac{53}{45}\)

\(\text{Vậy }x=\frac{53}{45}\)

\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)

\(\text{TH 1: Dấu chia}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)

\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)

\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)

\(\text{Vậy }\text{​​}x=\frac{-1}{5}\)

\(\text{TH 2:Dấu cộng}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{-3}{25}\)

\(\Leftrightarrow x=\frac{-3}{25}:2\)

\(\Leftrightarrow x=\frac{-3}{50}\)

\(\text{Vậy }x=\frac{-3}{50}\)

\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)

\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)

\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{5}{4}\)

\(\text{Vậy }x=\frac{5}{4}\)