\(\left(x+1\right)\left(y-2\right)=3\)

\(\Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm2\right\}\)

\(TH1:\hept{\begin{cases}x+1=1\\y-2=3\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=5\end{cases}}}\)             \(TH2:\hept{\begin{cases}x+1=-1\\y-2=-3\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)

\(TH3:\hept{\begin{cases}x+1=3\\y-2=1\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\)               \(TH4:\hept{\begin{cases}x+1=-3\\y-2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=1\end{cases}}}\)

Vậy ............................

b, Làm tương tự 

khó quá ko trả lời được

Lê Thị Phương Dung       

Bài trên mình làm sai nha

Để mình nghĩ lại :(

a) Ta có: |4x - 1| - x = 15

- Nếu \(4x-1\ge0\) \(\Rightarrow x\ge\frac{1}{4}\)

=> 4x - 1 - x = 15

=> 3x = 15 + 1

=> 3x = 16

=> x = \(\frac{16}{3}\) (thỏa mãn điều kiện)

- Nếu \(4x-1< 0\Rightarrow x< \frac{1}{4}\)

=> 1 - 4x - x = 15

=> -5x = 14

=> x = \(\frac{-14}{5}\) (thỏa mãn điều kiện)

Vậy x = \(\frac{16}{3}\) hoặc x = \(\frac{-14}{5}\)

Câu b hình như là đề sai rùi bạn ơi.

c) Ta có: 2x = 3y

\(\Rightarrow\) \(\frac{x}{3}=\frac{y}{2}\) \(\Rightarrow\) \(\frac{x}{21}=\frac{y}{14}\) (1)

5y = 7z

\(\Rightarrow\) \(\frac{y}{7}=\frac{z}{5}\) \(\Rightarrow\) \(\frac{y}{14}=\frac{z}{10}\) (2)

Từ (1) và (2) suy ra:

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\) \(\frac{x}{21}=2\) \(\Rightarrow\) \(x=21.2=42\)

\(\Rightarrow\) \(\frac{y}{14}=2\) \(\Rightarrow\) \(y=14.2=28\)

\(\Rightarrow\)\(\frac{z}{10}=2\) \(\Rightarrow\) \(z=10.2=20\)

Vậy x = 42; y = 28; z = 20

2, |2x-1|=(2x+3)

• 2x-1>0 => 2x-1= 2x+3

=> 2x-1-2x-3=0 => -4 =0 => ko có x thoả mãn

• 2x-1< 0 => -2x+1 = 2x+3

=> -2x +1 - 2x -3 =0 => -4x -2 =0 => -4x =2 => x=-1/2

Vậy x=-1/2

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

Để \(2x+2⋮2x-4\Leftrightarrow1+\frac{3}{x-2}\in Z\)

\(\Rightarrow x-2\inƯ\left(3\right)\) = { - 3; - 1; 1; 3 }

=> x = { - 1; 1 ; 3; 5 }

còn ai nữa ko tôi k hết cho mấy bạn

có 2x+2 ⋮ 2x-4                  (x ∈ Z)

=> (2x-4)+6 ⋮ 2x-4

Mà 2x-4 ⋮ 2x-4

=> 6 ⋮ 2x-4

=> 2x-4 ∈ Ư(6)={1;-1;2;-2;3;-3;6;-6}

Vậy x ∈ {3;1;5;-1}

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(x+x+x+x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=1-\frac{15}{16}\)

\(4x=\frac{1}{16}\)

\(x=\frac{1}{16}\div4\)

\(x=\frac{1}{64}\)

ta coi x để ra mội bên ta có 1/2 + 1/4 + 1/8 + 1/6 = 25/24

x + 25/24 = 1

            x = 1 - 25/24

            x = -1/24

mk nha

Nhân lên 16 lần ta có đề bài mới: 

Tìm x biết: \(\left(x+8\right)+\left(x+4\right)+\left(x+2\right)+\left(x+1\right)=16\)

Bài giải

\(x\times4+1+2+4+8=16\)

\(x\times4=1\)

Vậy \(x=\frac{1}{4}\)

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

Bài 1:

a) 8² . 2² = 64 . 4 = 256

b) 2⁵. 4 = 2⁵. 2²= 2⁵⁺² = 2⁷

c) 9⁴. 243 = 9⁴. 3⁵=(3²)⁴.3⁵=3⁸.3⁵=3⁸⁺⁵=3¹³

Bài 1 :

\(a.8^2.2^2=\left(8.2\right)^2=16^2\)

\(b.2^5.4=2^5.2^2=2^7\)

\(c.9^4.243=\left(3^2\right)^4.3^5=3^8.3^5=3^{13}\)

Bài 2:

a. ( đề sai =.= )

\(b.12.\left(x-1\right):3=43+2^3\)

\(\Leftrightarrow12.\left(x-1\right)=43+2^3.3\)

\(\Leftrightarrow12.\left(x-1\right)=67\)

\(\Leftrightarrow x-1=\frac{67}{12}\)

\(\Leftrightarrow x=\frac{79}{12}\)

tao đéo biết

a) \(\left|x\right|-\left|2x-3\right|=x-1\)

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

\(\left|2x-3\right|=\left|x\right|-x+1\)

* Với x > 0 thì :

\(2x-3=x-x+1\)

\(2x-3=1\)

\(2x=3+1\)

\(2x=4\)

\(x=4\text{ : }2\)

\(x=2\)

* Với x < 0 thì :

\(-\left(2x\right)-3=-x-x+1\)

\(-2x-3=-2x+1\)

\(-2x+2x=1+3\)

\(0\ne4\)

\(\Rightarrow\text{ }x=2\)

c) \(\left|x-1\right|+\left|\left(x-1\right)\left(x+1\right)\right|=0\)

Vì \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left|\left(x-1\right)\left(x+1\right)\right|\ge0\end{cases}}\) mà \(\left|x-1\right|+\left|\left(x-1\right)\left(x+1\right)\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x-1\right|=0\\\left|\left(x-1\right)\left(x+1\right)\right|=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x-1=0\\\left(x-1\right)\left(x+1\right)=0\end{cases}}\) \(\hept{\begin{cases}x=0+1\\x-1=0;x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=0+1=1;x=0-1=-1\end{cases}}\)

\(\Rightarrow\text{ }x=1\)

1.Tính hợp lí:

-5/9.3/11+(-3/18).3/11

=3/11.[(-5/9)+(-3/18)]

=3/11.-13/18

=-39/198

=13/66

2.Tìm x biết:

2/3+7/4:x=5/6

       7/4:x=5/6-2/3

       7/4:x=1/6

             x=7/4:1/6

             x=21/2      

bài 1

\(\frac{-5}{9}.\frac{3}{11}+\frac{-3}{18}.\frac{3}{11}\)

= \(\frac{3}{11}.\left(\frac{-5}{9}+\frac{-3}{18}\right)\)

= \(\frac{-13}{66}\)

bài 2

\(\frac{2}{3}+\frac{7}{4}:x=\frac{5}{6}\)

\(\frac{7}{4}:x=\frac{1}{6}\)

\(x=\frac{21}{2}\)