đây là sinh mà bn

\(a,x\left(1-2x\right)-2=\left(2x-3\right)\left(1-x\right)\\ \Leftrightarrow x-2x^2-2=2x-3-2x^2+3x\\ \Leftrightarrow2x-3-2x^2+3x-x+2x^2+2=0\)

\(\Leftrightarrow4x-1=0\\ \Leftrightarrow x=\dfrac{1}{4}\)

\(b,2x\left(x-2\right)+5x-10=0\\ \Leftrightarrow2x\left(x-2\right)+5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(A\left(x\right)=2x^2+2x+3\)

3) \(A\left(x\right)=3\)

khi đó: \(2x^2+2x+3=3\)

<=> \(x^2+x=0\)

<=> \(x\left(x+1\right)=0\)

<=> \(x=0\)

hoặc \(x=-1\)

A(x) = 3x² + x³ + 5x⁴ - x² - x³ - 5x⁴ + 2x + 3 

        = 2x² + 2x + 3

A(x) + B(x) = 2x - 7

<=> ( 2x² + 2x + 3 ) + B(x) = 2x - 7

B(x) = 2x - 7 - ( 2x² + 2x + 3 )

        = 2x - 7 - 2x² - 2x - 3

        = -2x² - 10

A(x) = 3 <=> 2x² + 2x + 3 = 3

              <=> x( 2x + 2 ) = 0

              <=> x = 0 hoặc 2x + 2 = 0

              <=> x = 0 hoặc x = -1 

3, A(x) = 2x² + 2x +3 

Ta có : A(x) = 3

\(\Leftrightarrow\)2x² + 2x + 3 = 3

\(\Leftrightarrow\)2x² +2x = 0

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

\(\Leftrightarrow\)2x = 0 hoặc x + 1 = 0

\(\Leftrightarrow\)x = 0 hoặc x = -1

Vậy x = 0 hoặc -1 khi A(x) = 3

a) Ta có : 2x = 3y => \(\frac{x}{3}=\frac{y}{2}\) 

7z = 5y => \(\frac{y}{7}=\frac{z}{5}\)

=> \(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)

+) \(\frac{x}{3}=\frac{y}{2}\)=> \(\frac{x}{21}=\frac{y}{14}\)

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

=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

=> \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

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

=> x = 2.21 = 42 , y = 2.14 = 28 , z = 2.10 = 20

b) Ta có : x : y : z = 3 : 5 : (-2) => \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\)

Đặt \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\\z=-2k\end{cases}}\)

=> 5x = 15k , y = 5k , 3z = -6k

=> 5x - y + 3z = 15k - 5k + (-6k)

=> -16 = 10k - 6k

=> -16 = 4k

=> k = -4

Với k = -4 thì x = 3.(-4) = -12 , y = 5.(-4) = -20 , z = (-2).(-4) = 8

Vậy : ....

a)2x+7=3x+10

\(\Rightarrow7-10=3x-2x\)

\(\Rightarrow-3=x\)

Vậy x=-3

b)Bạn tự làm nha

c)Bạn làm tương tự câu d nha

d)+)Ta có :\(x+1⋮x+1\left(1\right)\)

+)Theo bài ta có:\(x-3⋮x+1\left(2\right)\)

+)Từ (1) và (2)

\(\Rightarrow\left(x+1\right)-\left(x-3\right)⋮x+1\)

\(\Rightarrow x+1-x+3⋮x+1\)

\(\Rightarrow4⋮x+1\)

\(\Rightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1;-5;3\right\}\)

Vậy \(x\in\left\{-2;0;-3;1;-5;3\right\}\)

Chúc bn học tốt

A = 2x² - 6xy - 3xy - 6y - 2x² + 8xy + 6y

   = - xy

  = \(\frac{2}{3}\)\(x\)\(\frac{3}{4}\)

  = \(\frac{1}{2}\)

mk đang bận mấy câu kia tương tự nha

\(3x\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)

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

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)

\(=\frac{6}{7}\)

Tìm x

\(a,3x(2x+1)=0\)

\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)

\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)

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

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

\(\frac{4}{3}x=\frac{2}{3}-5\)

\(\frac{4}{3}x=\frac{-13}{3}\)

\(x=\frac{-13}{3}\div\frac{4}{3}\)

\(x=\frac{-13}{4}\)

Chúc ban học tốt

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}{7.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}=\frac{6}{7}\)

Tìm x

a)bạn tự làm vì nó rất dễ nha

b)\(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}.\left(2x+1\right)=5\)

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

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

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

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

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

\(\frac{-2}{3}x=6+\frac{2}{3}\)

\(\frac{-2}{3}x=\frac{20}{3}\)

\(x=\frac{20}{3}:\left(\frac{-2}{3}\right)\)

\(x=\frac{20}{3}.\left(\frac{3}{-2}\right)\)

x=-10

Vậy x=-10

Chúc bạn học tốt

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)