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19 tháng 6 2021

a) \(=x^2-49-x^2\) \(=-49\)

b) \(=25x^2-1-25x^2-1\) \(=-2\)

c) \(=16x^2-1-16x^2+8x-1\) \(=8x-2\)

d) \(=9x^2-30x+25-9x^2+25\) \(=50-30x\)

24 tháng 6 2018

\(a,5x^2-3x\left(x-2\right)\)

\(=5x^2-3x^2+6x\)

\(=2x^2+6x\)
\(b,3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c, Đề ko rõ Yang Yang

\(d,7x\left(x-5\right)+3\left(x-2\right)\)

\(=7x^2-35x+3x-6\)

\(=7x^2-32x-6\)

\(e,5-4x\left(x-2\right)+4x^2\)

\(=5-4x^2+8x+4x^2\)

\(=5+8x\)

\(f,4x\left(2x-3\right)-5x\left(x-2\right)\)

\(=8x^2-12x-5x^2+10x\)

\(=3x^2-2x\)

5 tháng 10 2019

rút gọn biểu thức

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

\(=4x^2-\left(x^2-5x+3x-15\right)+x\)

\(=4x^2-x^2+5x-3x+15+x\)

\(=3x^2+3x+15.\)

b) \(x.\left(x-5\right)-3x.\left(x+1\right)\)

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

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

\(=-2x^2-8x.\)

d) \(\left(x+3\right).\left(x-1\right)-\left(x-7\right).\left(x-6\right)\)

\(=x^2-x+3x-3-\left(x^2-6x-7x+42\right)\)

\(=x^2-x+3x-3-x^2+6x+7x-42\)

\(=15x-45.\)

Chúc bạn học tốt!

23 tháng 3 2020

Bài2: phân tích đa thức thành nhân tử 

\(a,x^2-y^2-2x+2y\)

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

\(b,x^3-5x^2+x-5\)

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

\(=\left(x+x-5\right)\left(x-x-5\right)\)

  \(c,x^2-2xy+y^2-9\)

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

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

chúc bạn học tốt !

24 tháng 3 2020

a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)

A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21

A = -76

b) B = 4x(3x - 2) - 3x(4x + 1)

B = 12x^2 - 8x - 12x^2 - 3x

B = -11x

c) C = (x + 3)(x - 2) - (x - 1)^2

C = x^2 + x - 6 - x^2 + 2x - 1

C = 3x - 7

14 tháng 7 2017

Câu a phần I sai. đề là :
a) A = -3x(x - 5 ) + 3(x2 - 4x ) - 3x + 10

13 tháng 6 2019

a) \(3y^2\left(2y-1\right)+y-y\left(1-y+y^2\right)-y^2+y \)

= \(6y^3-3y^2+y-y+y^2-y^3-y^2+y\)

= \(5y^3-3y^2+y\)

b)\(25x-4\left(3x-1\right)+\left(5-2x\right)7\)

= \(25x-12x+4+35-14x\)

= \(-x+39\)

c) \(11x-2\left(10x-1\right)-\left(4x-1\right)\left(-2\right)\)

= \(11x-\left(20x-2\right)-\left(-8x+2\right)\)

= \(11x-20x+2+8x-2\)

= \(-x\)

d) \(\left(\frac{1}{2x}\right)3-x\left(1-2x-\frac{1}{8x^2}\right)-x\left(x+\frac{1}{2}\right)\)

= \(\frac{3}{2x}-x+2x^2+\frac{x}{8x^2}-x^2-\frac{x}{2}\)

= \(\left(\frac{3}{2x}+\frac{1}{8x}-\frac{x}{2}\right)+x^2-x\)

= \(\left(\frac{12+1-4x^2}{8x}\right)+x^2-x\)

= \(\frac{13-4x^2}{8x}+\frac{8x^3}{8x}-\frac{8x^2}{8x}\)

= \(\frac{13-4x^2+8x^3-8x^2}{8x}\)

= \(\frac{8x^3-12x^2+13}{8x}\)

= x2 - \(\frac{3}{2}\)+\(\frac{13}{8x}\)

e) \(12\left(2-3x\right)+35x-\left(x+1\right)\left(-5\right)\)

= \(24-36x+35x-\left(-5x-5\right)\)

= \(24-36x+35x+5x+5\)

= 4x + 29

câu d:(-1/2x)3-x.(1-2x-1/8x2)-x.(x+1/2) nha

2 tháng 10 2017

\(1,3x-24y=3\left(x-8y\right)\)

\(2,6x^3y^2-12x^2y^2-3x^2y=3x^2y\left(2xy-4y-1\right)\)

\(3,7x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(7x-8\right)\)

...(tương tự)

\(10,5x-5y+x^2-xy=5\left(x-y\right)+x\left(x-y\right)=\left(x-y\right)\left(x+5\right)\)

\(11,x^2+2xy+y^2-16=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)

a) Ta có: 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)

b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

hay \(x=\frac{15}{8}\)

Vậy: \(x=\frac{15}{8}\)

c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)

\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)

\(\Leftrightarrow-6x^2-11x=0\)

\(\Leftrightarrow6x^2+11x=0\)

\(\Leftrightarrow x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)

d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

\(\Leftrightarrow14x^2=-18\)

\(14x^2\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)