2/3+5/8=16/24+15/24=31/24

4 và 2/5-2 và 2/3=22/5-7/3=66/15-35/15=31/15

 x + 2 và 1/5 =4 và 2/3

x + 11/5        =   14/3

x                   =   14/3-11/5

x                   =    37/15

x  :  3/2         =   4/15

x                   =    4/15*3/2

x                   =          2/5

Thanks Nguyễn Phương Anh nhìu nha!

phá ngoặc ra mà tính đi bn ơi

a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)

b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)

\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)

c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)

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

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

\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)

d. \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7=x\)

Vậy x = 7

e. \(\left(x-1\right)-\left(2x-1\right)=9\)

\(\Leftrightarrow x-1-2x+1=9\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

g. \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

chào bn mik đến từ năm 2022

a: \(\left(x-1\right)^3+27\)

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

\(=\left(x+2\right)\left(x^2+x+1\right)\)

b: \(\left(x-2\right)^3-8\)

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

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

1) Ta có: |x+3| \(\ge\)0; |2x+y-4| \(\ge\)0

\(\Rightarrow\) |x + 3| + |2x + y - 4| \(\ge\) 0

Dấu = xảy ra khi x+3=0 và 2x+y-4 = 0 \(\Rightarrow\)x=-3; y=10

1)  |x + 3| + |2x + y - 4| = 0

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-6+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=10\end{cases}}\)

2) Ta có:  |x + 3| + |x + y - 6| + |7 - 1| = 0

\(\Leftrightarrow\) |x + 3| + |x + y - 6| + 6= 0

\(\Leftrightarrow\)|x + 3| + |x + y - 6| = -6

dễ mak

Bài 1 :

1) a² - 4 + y ( a - 2 )

= ( a + 2 ) ( a - 2 ) + y ( a - 2 )

= ( a - 2 ) ( a + 2 + y )

2) ( x - 2 )² - 9y²

= ( x - 2 - 3y ) ( x - 2 + 3y )

Bài 2 :

1) 3 ( x + 4 ) - 2x = 5

=> 3x + 12 - 2x = 5

=> x + 12 = 5

=> x = 5 - 12 = - 7

Vậy x = - 7

2) x ( x - 2 ) - x² - 6 = 0

=> x² - 2x - x² - 6 = 0

=> - 2x - 6 = 0

=> 2x = - 6

=> x = \(-\frac{6}{2}=3\)

Vậy x = 3

3 ) x² - 3x = 0

=> x ( x - 3 ) = 0

=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x\in\left\{0;3\right\}\)

4) 5 - 3 ( x - 6 ) = 4

=> 5 - 3x + 18 = 4

=> 3x = 5 + 18 - 4

=> 3x = 19

=> x = \(\frac{19}{3}\)

Vậy \(x=\frac{19}{3}\)

Thanks nhìu nha😍😍😍😍