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

\(=x^2\left(x-1\right)-2x\cdot\left(x-1\right)\)

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

5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)

\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)

\(=\left(x^2+y\right)\left(y^2-z\right)\)

7: \(5\left(x+y\right)^2+15\left(x+y\right)\)

\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)

=5(x+y)(x+y+3)

9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\left(7x+y-4\right)\)

11: \(\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)

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

=(y-2)(x+1-y+2)

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

2: \(5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)

4: 3x(x-5y)-2y(5y-x)

=3x(x-5y)+2y(x-5y)

=(x-5y)(3x+2y)

6: b(a-c)+5c-5a

=b(a-c)-5(a-c)

=(a-c)(b-5)

8: 9x(x-y)-10(y-x)^2

=9x(x-y)-10(x-y)^2

=(x-y)(9x-10x+10y)

=(x-y)(-x+10y)

10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)

\(=\left(a-b\right)^2+\left(a+b\right)\left(a-b\right)\)

=(a-b)(a-b+a+b)

=2a(a-b)

12: 2x(x-3)+y(x-3)+(3-x)

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

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

a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)

Câu a sai đề rồi, mình sửa luôn nhé. \(\)

A= 3x.(x-5y) + (y-5x) . (-3y) - 3.(x²-y²) -1

A= \(3x^2-15xy-3y^2+15xy-3x^2+3y^2-1\)

A= \(-1\)

Vậy biểu thức trên không phụ thuộc vào biến x.

B= \(\left(3x-5\right).\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

B= \(6x^2+33x-10x-55-6x^2-14x-9x-21\)

B= \(-76\)

Vậy biểu thức trên không phụ thuộc vào biến x.

Chúc bạn học tốt :))

\(B=\left(3x-5\right)\left(2x+11\right)-\left(2x-3\right)\left(3x+7\right)\)\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)\(=-76\)

Vậy Biểu thức trên không phụ thuộc vào x

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

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

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

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