a) (12x-5)(4x-1)+(3x-7)(1-16x)

= (48x^2 - 12x - 20x + 5) + (3x - 48x^2 - 7 + 112x)

= 48x^2 - 12x - 20x + 5 +3x - 48x^2 -7 + 112x

= 83x-2

những phần sau bạn cứ làm tương tự theo cách nhân đa thức với đa thức và phá ngiawcj là ra nha :0))

sao ko làm hết ra

3x - 7 = 2x + 5

3x - 2x = 5 + 7

x = 12

|3x - 2| = 7

\(\Rightarrow\left\{\begin{matrix}3x-2=7\\-\left(3x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x=9\\-3x+2=7\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=3\\-3x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=3\\x=-\frac{5}{3}\end{matrix}\right.\)

4 - |x - 2| = -3

|x - 2| = 4 - (-3)

|x - 2| = 7

\(\Rightarrow\left\{\begin{matrix}x-2=7\\-\left(x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\-x+2=7\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=9\\-x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

|2x - 3| = x - 1

\(\Rightarrow\left\{\begin{matrix}2x-3=x-1\\-\left(2x-3\right)=x-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x-x=-1+3\\-2x+3=x-1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=2\\-2x-x=-1-3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-3x=-4\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=2\\x=\frac{4}{3}\end{matrix}\right.\)

|3x + 1| = x + 3

\(\Rightarrow\left\{\begin{matrix}3x+1=x+3\\-\left(3x+1\right)=x+3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x-x=3-1\\-3x-1=x+3\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}2x=2\\-3x-x=3+1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\-4x=4\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a)(2x+1)(3x+2)(x-5)=0

Th1:2x+1=0

=>2x=-1

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

Th2:3x+2=0

=>3x=-2

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

Th2:x-5=0

=>x=5

b)<=>4x-7=-3x-6

=>4x-(-3x)=7-6

=>7x=1

=>x=\(\frac{1}{7}\)

còn lại bạn tự làm

a)2x+1=0

3x+2=0

x-5=0

(giải ra là dược)

b)4x-7=-3(x+2)

4x-7=-3x-6

4x+3x=-6+7

7x=1

x=1/7

câu c) viết lại đề

câu d) \(\frac{5x+7}{x-1}=3x+2\)

=>5x+7=(3x+2)(x-1)

=>5x+7=3x(x-1)+2(x-1)

=>5x+7=3x²-3x+2x-2

=>5x+7=3x²-x-2=3x²-(x+2)

=>3x²=5x+7+(x+2)=6x+9

=>\(x^2=\frac{6x+9}{3}=\frac{6x}{3}+\frac{9}{3}=2x+3\)

=>x²-(2x+3)=0

=>x²-2x-3=0

=>x²-3x+x-3=0

=>x(x-3)+(x-3)=0

=>(x+1)(x-3)=0

=>x=-1 hoặc x =3

Thử lại thấy đúng,vậy x \(\in\) {-1;3}

Gía trị biểu thức không phụ thuộc vào biến nghĩa là với mọi x, biểu thức đó có giá trị là 1 số thực.Ta có :

A = 2x(x - 1) - x(2x + 1) - (3 - 3x) = 2x² - 2x - 2x² - x - 3 + 3x = (2x² - 2x²) + (3x - 2x - x) - 3 = -3

B = 2x(x - 3) - (2x - 2)(x - 2) = 2x² - 6x - 2x² + 4x + 2x - 4 = (2x² - 2x²) + (4x + 2x - 6x) - 4 = -4

C = (3x - 5)(2x + 11) - (2x + 3)(3x + 7) = 6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21 = (6x² - 6x²) + (33x - 10x - 14x - 9x) - 55 - 21 = -76      = D = (2x + 11)(3x - 5) - (2x + 3)(3x + 7)

Vậy với mọi x , (A,B,C,D) = (-3;-4;-76;-76) => đpcm 

D = 

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)

=>7/6x=2/3

hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)

hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)